Genomics Mc - All - Random
- 1.
Which of the following statements about an organism's genome is FALSE?
- A.
The genome contains the genetic information to construct and maintain a living organism.
- B.
The genomes of cellular organisms are composed of DNA.
- C.
The genome is able to express its own information without the activity of enzymes and proteins
- D.
Eukaryotic genomes are composed of both nuclear and mitochondrial DNA.
Correct Answer
C. The genome is able to express its own information without the activity of enzymes and proteinsExplanation
The given statement that the genome is able to express its own information without the activity of enzymes and proteins is FALSE. Enzymes and proteins play a crucial role in gene expression by facilitating the transcription and translation processes. They help in the regulation and activation of genes, as well as the modification and processing of RNA molecules. Without the activity of enzymes and proteins, the genetic information in the genome would not be able to be expressed and translated into functional proteins.Rate this question:
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- 2.
Somatic cells are those that:
- A.
Contain a haploid set of chromosomes.
- B.
Give rise to the gametes.
- C.
Lack mitochondria.
- D.
Contain a diploid set of chromosomes and make up the majority of human cells.
Correct Answer
D. Contain a diploid set of chromosomes and make up the majority of human cells.Explanation
Somatic cells are the cells that make up the majority of human cells and contain a diploid set of chromosomes. This means that they have two sets of chromosomes, one from each parent. Somatic cells are not involved in the production of gametes (sex cells) and do not give rise to them. They also contain mitochondria, which are the powerhouses of the cell responsible for generating energy.Rate this question:
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- 3.
The flow of genetic information in cells is which of the following?
- A.
DNA is transcribed into RNA, which is then translated into protein.
- B.
DNA is translated into protein, which is then transcribed into RNA.
- C.
RNA is transcribed into DNA, which is then translated into protein.
- D.
Proteins are translated into RNA, which is then transcribed into DNA
Correct Answer
A. DNA is transcribed into RNA, which is then translated into protein.Explanation
The flow of genetic information in cells is that DNA is first transcribed into RNA through a process called transcription. This RNA molecule, known as messenger RNA (mRNA), then leaves the nucleus and enters the cytoplasm where it is translated into protein by ribosomes. This process is called translation. Therefore, the correct answer is that DNA is transcribed into RNA, which is then translated into protein.Rate this question:
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- 4.
In the early tvventieth century it was thought that proteins might carry genetic information. This reasoning was due to which of the following?
- A.
Chromosomes are composed of approximately equal amounts of protein and DNA.
- B.
Proteins were known to be composed of 20 distinct amino acids whereas DNA is composed of only 4 nucleotides.
- C.
Different proteins were known to have unique sequences, whereas it was thought that all DNA molecules have the same sequence.
- D.
All of the above.
Correct Answer
B. Proteins were known to be composed of 20 distinct amino acids whereas DNA is composed of only 4 nucleotides.Explanation
In the early twentieth century, it was believed that proteins might carry genetic information because proteins were known to be composed of 20 distinct amino acids, whereas DNA is composed of only 4 nucleotides. This difference in complexity led scientists to speculate that proteins, with their larger variety of building blocks, might be better suited to carry genetic information. This reasoning was supported by the fact that different proteins were known to have unique sequences, while it was believed that all DNA molecules have the same sequence.Rate this question:
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- 5.
Which type of bonds link the individual nucleotides together in DNA?
- A.
Glycosidic.
- B.
Peptide
- C.
Phosphodiester.
- D.
Electrostatic.
Correct Answer
C. pHospHodiester.Explanation
Phosphodiester bonds link the individual nucleotides together in DNA. These bonds form between the phosphate group of one nucleotide and the sugar group of another nucleotide, creating a strong backbone for the DNA molecule. This bond is important for maintaining the stability and integrity of the DNA structure. Glycosidic bonds are found in carbohydrates, peptide bonds are found in proteins, and electrostatic bonds are formed due to the attraction between positively and negatively charged particles.Rate this question:
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- 6.
In solving the structure of DNA, Watson and Crick actively used which of the following techniques?
- A.
Model building of DNA molecules to ensure that the atoms were correctly positioned
- B.
X-ray crystallography of DNA
- C.
Chromatographic studies to determine the relative composition of nucleotides from various sources
- D.
Genetic studies that demonstrated that DNA is the genetic material.
Correct Answer
A. Model building of DNA molecules to ensure that the atoms were correctly positionedExplanation
Watson and Crick actively used model building of DNA molecules to ensure that the atoms were correctly positioned in solving the structure of DNA. This technique allowed them to visualize and understand the three-dimensional structure of DNA, including the double helix shape and the specific arrangement of nucleotides. By physically manipulating models of DNA molecules, they were able to propose a structure that explained the observed X-ray diffraction patterns and provided insights into the replication and function of DNA.Rate this question:
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- 7.
Erwin Chargaff studied DNA from various organisms and demonstrated that:
- A.
DNA is the genetic material
- B.
RNA is transcribed from DNA.
- C.
The amount of adenine in a given organism is equal to the amount of thymine (and guanine to cytosine).
- D.
The double helix is held together by hydrogen bonding between the bases
Correct Answer
C. The amount of adenine in a given organism is equal to the amount of thymine (and guanine to cytosine).Explanation
Erwin Chargaff's studies on DNA from various organisms showed that the amount of adenine in a given organism is equal to the amount of thymine, and the amount of guanine is equal to the amount of cytosine. This observation is known as Chargaff's rules and is a fundamental principle of DNA structure. It provides evidence for the complementary base pairing in DNA, where adenine always pairs with thymine and guanine always pairs with cytosine. This balanced pairing is crucial for DNA replication and the accurate transmission of genetic information.Rate this question:
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- 8.
The transcriptome of a cell is defined as:
- A.
All of the RNA molecules present in a cell.
- B.
The protein-coding RNA molecules present in a cell.
- C.
The ribosomal RNA molecules present in a cell.
- D.
The transfer RNA molecules present in a cell.
Correct Answer
A. All of the RNA molecules present in a cell.Explanation
The transcriptome of a cell refers to all of the RNA molecules present in the cell. This includes not only protein-coding RNA molecules, but also ribosomal RNA (rRNA) and transfer RNA (tRNA) molecules. The transcriptome provides a snapshot of the genes that are actively being expressed in a cell at a given time, and it plays a crucial role in determining the cell's phenotype and function.Rate this question:
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- 9.
How do DNA-dependent RNA polymerases carry out RNA synthesis?
- A.
They use DNA as a template for the polymerization of ribonucleotides.
- B.
They use proteins as a template for the polymerization of ribonucleotides
- C.
They use RNA as a template for the polymerization of ribonucleotides
- D.
They require no template for the polymerization of ribonucleotides
Correct Answer
A. They use DNA as a template for the polymerization of ribonucleotides.Explanation
DNA-dependent RNA polymerases carry out RNA synthesis by using DNA as a template for the polymerization of ribonucleotides. This means that they use the sequence of DNA as a guide to create a complementary RNA strand, with each ribonucleotide being added to the growing RNA molecule based on the corresponding nucleotide in the DNA template. This process is essential for the transcription of genes and the production of RNA molecules in cells.Rate this question:
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- 10.
Which type of functional RNA is a primary component of the structures required for protein synthesis?
- A.
Messenger RNA.
- B.
Ribosomal RNA.
- C.
Small nuclear RNA.
- D.
Transfer RNA.
Correct Answer
D. Transfer RNA.Explanation
Transfer RNA (tRNA) is a type of functional RNA that plays a crucial role in protein synthesis. It carries amino acids to the ribosome, where they are assembled into a polypeptide chain according to the sequence of codons on the messenger RNA (mRNA). tRNA molecules have a specific anticodon sequence that is complementary to the codon on the mRNA, allowing them to recognize and bind to the appropriate amino acid. Therefore, tRNA is a primary component of the structures required for protein synthesis. Ribosomal RNA (rRNA) is also involved in protein synthesis as a structural component of the ribosome, but it is not the primary component. Small nuclear RNA (snRNA) is involved in RNA processing, while messenger RNA (mRNA) carries the genetic information from DNA to the ribosome.Rate this question:
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- 11.
The proteome of a cell is defined as:
- A.
All of the proteins that a cell is capable of synthesizing.
- B.
All of the proteins present in a cell over the cell's lifetime
- C.
All of the proteins present in a cell at a given moment.
- D.
All of the proteins that are actively being synthesized in a cell at a given moment.
Correct Answer
C. All of the proteins present in a cell at a given moment.Explanation
The proteome of a cell refers to all the proteins that are present in a cell at a specific point in time. This includes the proteins that have been synthesized and are currently present in the cell. It does not include all the proteins that the cell is capable of synthesizing or all the proteins that have been synthesized over the cell's lifetime. The proteome is a dynamic entity that can change over time as proteins are synthesized, degraded, or modified within the cell.Rate this question:
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- 12.
Which level of protein structure describes the folded conformation of a multisubunit protein?
- A.
Primary structure.
- B.
Secondary structure.
- C.
Tertiary structure.
- D.
Quaternary structure.
Correct Answer
D. Quaternary structure.Explanation
The quaternary structure of a protein refers to the arrangement of multiple subunits in a protein complex. It describes how these subunits come together and interact to form a functional protein. This level of protein structure is important for proteins that consist of multiple subunits, such as enzymes, antibodies, and hemoglobin. The primary structure refers to the linear sequence of amino acids, the secondary structure refers to local folding patterns such as alpha helices and beta sheets, and the tertiary structure refers to the overall 3D folding of a single polypeptide chain.Rate this question:
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- 13.
Which type of covalent bond is important for linking cysteine residues located at various places in a polypeptide?
- A.
Disulfide bridge.
- B.
Hydrogen bond.
- C.
Peptide bond.
- D.
Phosphodiester bond.
Correct Answer
A. Disulfide bridge.Explanation
Disulfide bridges are important for linking cysteine residues located at various places in a polypeptide. These bridges form when two cysteine residues come close together and the sulfur atoms in their side chains bond to each other, creating a covalent bond. This bond helps stabilize the tertiary structure of the protein by creating a bridge between different regions of the polypeptide chain. Hydrogen bonds are important for stabilizing secondary structures like alpha helices and beta sheets. Peptide bonds link amino acids together to form the polypeptide chain. Phosphodiester bonds are found in nucleic acids, not proteins.Rate this question:
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- 14.
Most of the abundant proteins in a cell are thought to be housekeeping proteins. What is their function?
- A.
They are responsible for the specific functions of individual cell types.
- B.
They are responsible for regulating genome expression in cells
- C.
They are responsible for removing waste materials from cells.
- D.
They are responsible for the general biochemical activities that occur in all cells.
Correct Answer
D. They are responsible for the general biochemical activities that occur in all cells.Explanation
Housekeeping proteins are responsible for the general biochemical activities that occur in all cells. These proteins perform essential functions that are required for the basic maintenance and survival of the cell, such as energy production, metabolism, and protein synthesis. They are involved in fundamental cellular processes and are present in all cell types, regardless of their specific functions. Housekeeping proteins ensure the proper functioning of cellular processes and maintain the overall homeostasis of the cell.Rate this question:
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- 15.
The degeneracy of the genetic code refers to which of the following?
- A.
Each codon can specify more than one amino acid.
- B.
Most amino acids have more than one codon.
- C.
There are several initiation codons.
- D.
The stop codons can also code for amino acids.
Correct Answer
B. Most amino acids have more than one codon.Explanation
The degeneracy of the genetic code refers to the fact that most amino acids can be encoded by more than one codon. This means that different codons can specify the same amino acid. For example, the amino acid leucine can be encoded by six different codons (UUA, UUG, CUU, CUC, CUA, CUG). This redundancy in the genetic code provides a level of protection against mutations, as a change in one nucleotide may not necessarily result in a change in the encoded amino acid.Rate this question:
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- 16.
Which of the following is NOT a biological function of proteins?
- A.
Biological catalysis.
- B.
Regulation of cellu lar processes.
- C.
Carrying genetic information.
- D.
Transport of molecules in multicellular organisms.
Correct Answer
C. Carrying genetic information.Explanation
Proteins play a crucial role in various biological functions. They act as catalysts in biological reactions, regulate cellular processes, and facilitate the transport of molecules in multicellular organisms. However, carrying genetic information is not a biological function of proteins. Genetic information is primarily carried by nucleic acids, specifically DNA and RNA. Proteins, on the other hand, are involved in the expression and regulation of genetic information but do not directly carry it.Rate this question:
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- 17.
Which of the following enzymes are used to degrade DNA molecules?
- A.
DNA polymerases.
- B.
Nucleases.
- C.
Ligases.
- D.
Kinases.
Correct Answer
B. Nucleases.Explanation
Nucleases are enzymes that are used to degrade DNA molecules. They break down the DNA by cleaving the phosphodiester bonds that hold the nucleotides together. Nucleases can be specific to either single-stranded or double-stranded DNA, and they play a crucial role in various biological processes such as DNA repair, DNA replication, and gene regulation. DNA polymerases, on the other hand, are responsible for synthesizing new DNA strands during replication and repair. Ligases are involved in joining DNA fragments together, and kinases are enzymes that phosphorylate molecules, but they are not directly involved in degrading DNA molecules.Rate this question:
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- 18.
Why does a template-dependent DNA polymerase require a primer to initiate DNA synthesis?
- A.
These polymerases require a 5'-phosphate group to add a new nucleotide
- B.
These polymerases require a 3'-hydroxyl group to and add a new nucleotide
- C.
The primer is required for the DNA polymerase to bind to the template DNA.
- D.
The primer is hydrolyzed to provide the energy I join required for DNA synthesis
Correct Answer
B. These polymerases require a 3'-hydroxyl group to and add a new nucleotideExplanation
A template-dependent DNA polymerase requires a primer to initiate DNA synthesis because it requires a 3'-hydroxyl group to add a new nucleotide. The primer provides this 3'-hydroxyl group, which serves as the starting point for the addition of nucleotides during DNA synthesis. Without a primer, the DNA polymerase would not have a free 3'-hydroxyl group to which it can add nucleotides, and therefore, DNA synthesis cannot be initiated.Rate this question:
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- 19.
The function of the 3'-->5' exonuclease activity of a DNA polymerase is to:
- A.
Remove the 5' end of the polynucleotide strand that is attached to the template strand that is being copied.
- B.
Remove damaged nucleotides from the template strand during DNA synthesis.
- C.
Remove nucleotides from the ends of DNA molecules to ensure the generation of blunt ends.
- D.
Remove incorrect nucleotides from the newly synthesized strand of DNA.
Correct Answer
D. Remove incorrect nucleotides from the newly synthesized strand of DNA.Explanation
The 3'-->5' exonuclease activity of a DNA polymerase is responsible for removing incorrect nucleotides from the newly synthesized strand of DNA. This activity allows the DNA polymerase to proofread the newly synthesized DNA strand and correct any mistakes that may have occurred during replication. By removing the incorrect nucleotides, the DNA polymerase ensures the accuracy of the DNA sequence.Rate this question:
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- 20.
The Klenow polymerase version of E. coli DNA polymerase I is useful for research as it lacks the 5'-3' exonuclease activity. This is useful as the 5'--3' exonuclease activity:
- A.
Is more active than the polymerase activity.
- B.
Will prevent the incorporation of "radioactive or fluorescent labels into the DNA.
- C.
May interfere with some research applications by shortening the 5' ends of the DNA molecules.
- D.
Prevents the polymerase from detecting errors in the incorporation of new nucleotides.
Correct Answer
C. May interfere with some research applications by shortening the 5' ends of the DNA molecules.Explanation
The 5'-3' exonuclease activity of E. coli DNA polymerase I can shorten the 5' ends of DNA molecules. This can interfere with certain research applications where the length of the DNA molecule is important or needs to be preserved.Rate this question:
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- 21.
A temperature of 75°C will terminate DNA synthesis by E. coli DNA polymerase I. This is because:
- A.
E. coli DNA polymerase I is denatured at this temperature.
- B.
The DNA is denatured at this temperature.
- C.
The primers are denatured at this temperature.
- D.
Tile temperature is too high for enzymatic reactions to occur.
Correct Answer
A. E. coli DNA polymerase I is denatured at this temperature.Explanation
At a temperature of 75°C, E. coli DNA polymerase I is denatured. Denaturation refers to the disruption of the protein's structure, leading to the loss of its biological activity. In this case, the high temperature causes the DNA polymerase I enzyme to lose its shape and functionality, rendering it unable to carry out DNA synthesis. Consequently, DNA synthesis is terminated.Rate this question:
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- 22.
Which of the following statements accurately describes reverse transcriptases?
- A.
They are present in all viruses and are RNA dependent DNA polymerases.
- B.
They are present in all RNA viruses and are DNA dependent RNA polymerases.
- C.
They are present in retroviruses and are RNA dependent DNA polymerases.
- D.
They are present in all viruses and are template independent DNA polymerases.
Correct Answer
C. They are present in retroviruses and are RNA dependent DNA polymerases.Explanation
Reverse transcriptases are enzymes that are present in retroviruses and are responsible for the synthesis of DNA from RNA templates. They catalyze the reverse transcription process, where RNA is used as a template to generate complementary DNA (cDNA). This cDNA can then be integrated into the host genome, allowing the retrovirus to replicate and persist within the host cell. Therefore, the statement accurately describes reverse transcriptases.Rate this question:
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- 23.
All three types of restriction enzyme bind to DNA molecules at specific sequences; however. the type II enzymes are favored for research for which of the following reasons?
- A.
Type II enzymes cut the DNA at a specific site.
- B.
Type II enzymes always cut the DNA to yield blunt ended molecules.
- C.
Type II enzymes always cut the DNA to yield sticky ended molecules.
- D.
Type II enzymes are the only restriction enzymes to cleave double-stranded DNA.
Correct Answer
A. Type II enzymes cut the DNA at a specific site.Explanation
Type II enzymes are favored for research because they cut the DNA at a specific site. This specificity allows researchers to target and manipulate specific regions of the DNA molecule. This is important in various molecular biology techniques such as DNA cloning, gene editing, and DNA sequencing. By cutting the DNA at specific sites, researchers can precisely insert or remove genetic material, study gene function, and analyze DNA sequences.Rate this question:
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- 24.
Which technique is used to resolve the different sizes of DNA fragments following a restriction enzyme digest?
- A.
DNA sequencing.
- B.
Gel electrophoresis.
- C.
Gene cloning.
- D.
PCR
Correct Answer
B. Gel electropHoresis.Explanation
Gel electrophoresis is a technique used to separate DNA fragments based on their size and charge. In this technique, DNA samples are loaded into wells of a gel matrix and an electric current is applied. The negatively charged DNA molecules move towards the positive electrode, and smaller fragments migrate faster than larger ones. As a result, the DNA fragments are separated into distinct bands based on their size. Gel electrophoresis is commonly used in molecular biology to analyze and compare DNA samples, including those produced by restriction enzyme digestion. Therefore, it is the appropriate technique to resolve the different sizes of DNA fragments following a restriction enzyme digest.Rate this question:
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- 25.
DNA ligase synthesizes which type of bond?
- A.
The hydrogen bonds between bases.
- B.
The phosphodiester bonds between nucleotides.
- C.
The bonds between the bases and deoxyribose sugars.
- D.
The peptide bonds between amino acids.
Correct Answer
B. The pHospHodiester bonds between nucleotides.Explanation
DNA ligase is an enzyme that plays a crucial role in DNA replication and repair. It catalyzes the formation of phosphodiester bonds between adjacent nucleotides, sealing the gaps in the DNA backbone. This process is essential for joining the Okazaki fragments during lagging strand synthesis and repairing DNA damage. Therefore, the correct answer is that DNA ligase synthesizes phosphodiester bonds between nucleotides.Rate this question:
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- 26.
Which of the following polymerases does not require a template?
- A.
DNA polymerase I.
- B.
Sequenase.
- C.
Reverse transcriptase.
- D.
Terminal deoxynucleotidyl transferase.
Correct Answer
D. Terminal deoxynucleotidyl transferase.Explanation
Terminal deoxynucleotidyl transferase (TdT) is the correct answer because it is an enzyme that adds nucleotides to the 3' end of DNA strands in a template-independent manner. Unlike other polymerases, TdT does not require a template strand to synthesize DNA. It is commonly used in laboratory techniques such as DNA sequencing and gene cloning, where the addition of non-templated nucleotides is desired.Rate this question:
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- 27.
E. coli cells take up plasmid DNA in laboratory experiments by which of the following methods?
- A.
Conjugation.
- B.
Electrophoresis.
- C.
Transduction.
- D.
Transformation.
Correct Answer
D. Transformation.Explanation
In laboratory experiments, E. coli cells take up plasmid DNA through a process called transformation. Transformation involves the uptake of exogenous DNA by the bacterial cells, which can then be incorporated into their own genetic material. This process is commonly used in genetic engineering and molecular biology experiments to introduce specific genes or DNA sequences into bacterial cells for various purposes such as gene expression studies or production of recombinant proteins. Conjugation involves the transfer of genetic material between bacterial cells through direct cell-to-cell contact, electrophoresis is a technique used to separate DNA fragments based on their size and charge, and transduction involves the transfer of DNA between bacterial cells through a viral vector.Rate this question:
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- 28.
What is a genomic library?
- A.
A collection of recombinant molecules with inserts that contain all of the genes of an organism.
- B.
A collection of recombinant molecules with inserts that contain all of an organism's genome.
- C.
A collection of recombinant molecules that express all of the genes of an organism.
- D.
A collection of recombinant molecules that have been sequenced.
Correct Answer
B. A collection of recombinant molecules with inserts that contain all of an organism's genome.Explanation
A genomic library is a collection of recombinant molecules with inserts that contain all of an organism's genome. This means that the library contains fragments of DNA from the entire genome of the organism, allowing researchers to study and analyze the genetic information of that organism. The other options are incorrect because they either do not include the entire genome or do not specifically refer to the genetic material of the organism.Rate this question:
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- 29.
Which of the following types of vector would be most suitable for introducing DNA into a human cell?
- A.
Plasmid.
- B.
Bacteriophage.
- C.
Cosmid.
- D.
Adenovirus.
Correct Answer
D. Adenovirus.Explanation
Adenovirus would be the most suitable vector for introducing DNA into a human cell because it has the ability to efficiently infect a wide range of cell types, including human cells. Adenoviruses are also capable of carrying larger DNA fragments compared to other vectors, making them ideal for delivering larger genes or gene clusters. Additionally, adenoviruses have a high transduction efficiency, meaning they can effectively deliver the desired DNA into the target cells.Rate this question:
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- 30.
Which of the following is NOT used to introduce recombinant DNA molecules into plants?
- A.
Biolistics.
- B.
Cosmids.
- C.
Ti plasmid.
- D.
Viruses.
Correct Answer
B. Cosmids.Explanation
Cosmids are not used to introduce recombinant DNA molecules into plants. Cosmids are hybrid plasmids that are used for cloning large fragments of DNA in bacteria, but they are not commonly used in plant transformation techniques. Biolistics, also known as particle bombardment, involves shooting DNA-coated particles into plant cells. Ti plasmids are used in Agrobacterium-mediated transformation, where the bacteria transfer the DNA into the plant cells. Viruses can also be used as vectors to introduce recombinant DNA into plants.Rate this question:
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- 31.
PCR is advantageous to gene cloning for all of the following reasons except:
- A.
PCR does not require that the sequence of the gene be known.
- B.
PCR is a very rapid technique for the isolation of a gene.
- C.
PCR requires very small amounts of starting DNA compared to gene cloning.
- D.
PCR is very useful for mapping DNA markers.
Correct Answer
A. PCR does not require that the sequence of the gene be known.Explanation
PCR is advantageous to gene cloning for all of the following reasons except that PCR does not require that the sequence of the gene be known. PCR is a very rapid technique for the isolation of a gene, requiring very small amounts of starting DNA compared to gene cloning. Additionally, PCR is very useful for mapping DNA markers. However, the lack of requirement for the known gene sequence is not an advantage of PCR over gene cloning.Rate this question:
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- 32.
A major problem with the computational assembly of DNA sequences of complex eukaryotic genomes is the presence of:
- A.
Multiple chromosomes.
- B.
Mitochondrial DNA.
- C.
Introns within the genome.
- D.
Repetitive sequences.
Correct Answer
D. Repetitive sequences.Explanation
The presence of repetitive sequences is a major problem in the computational assembly of DNA sequences of complex eukaryotic genomes. These repetitive sequences can occur in multiple copies throughout the genome, making it difficult to accurately assemble the DNA sequence. The repetitive nature of these sequences can cause errors in the assembly process, leading to gaps or misalignments in the final sequence. Therefore, addressing the issue of repetitive sequences is crucial in accurately assembling the DNA sequences of complex eukaryotic genomes.Rate this question:
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- 33.
The first genetic maps used genes as markers because:
- A.
The locations of genes on chromosomes could be observed by staining the DNA with dyes.
- B.
Phenotypes specified by genes could be identified visually and their inheritance patterns studied.
- C.
Individual genes specifying easily identifiable phenotypic traits were easily cloned
- D.
Single nucleotide polymorph isms were used to identify point mutations that resulted in clearly observable phenotypic differences
Correct Answer
B. pHenotypes specified by genes could be identified visually and their inheritance patterns studied.Explanation
Genes were used as markers in the first genetic maps because phenotypes specified by genes could be visually identified and their inheritance patterns studied. This allowed researchers to track the transmission of specific traits from one generation to another, providing valuable insights into the inheritance patterns of genes. By observing the phenotypic traits associated with certain genes, researchers could map their locations on chromosomes and understand how they were passed on to offspring. This approach laid the foundation for the study of genetics and the development of genetic maps.Rate this question:
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- 34.
Which of the following is NOT a reason why biochemical phenotypes were commonly used to create human genetic maps?
- A.
Humans have no visual characteristics that are useful for genetic mapping.
- B.
There are biochemical phenotypes that are easily screened by blood typing.
- C.
Some easily characterized biochemical phenotypes are specified by genes with very large numbers of alleles.
- D.
It is unethical to perform controlled breeding experiments with humans.
Correct Answer
A. Humans have no visual characteristics that are useful for genetic mapping.Explanation
Biochemical phenotypes were commonly used to create human genetic maps because there are biochemical phenotypes that are easily screened by blood typing, and some easily characterized biochemical phenotypes are specified by genes with very large numbers of alleles. Additionally, it is unethical to perform controlled breeding experiments with humans.Rate this question:
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- 35.
Eukaryotlc genomes are mapped using DNA markers in addition to genes because:
- A.
DNA markers do not require the presence of two or more alleles for mapping.
- B.
Gene maps may not cover large regions of the genome.
- C.
Most genes contain multiple alleles that can be easily mapped.
- D.
DNA markers are less variable than genetic markers.
Correct Answer
B. Gene maps may not cover large regions of the genome.Explanation
Gene maps may not cover large regions of the genome. This is because gene maps are based on the location of specific genes, and not all genes are known or have been mapped. Additionally, gene maps may not accurately represent the entire genome, as there may be regions that are difficult to sequence or analyze. DNA markers, on the other hand, can be used to map the entire genome as they are based on specific DNA sequences that are spread throughout the genome. This allows for a more comprehensive and accurate mapping of the eukaryotic genome.Rate this question:
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- 36.
Microsatellites are used more commonly than minisatellites as DNA markers because:
- A.
Minisatellites are present in too many locations within the genome.
- B.
Restriction enzymes can be used to type microsatellites but not minisatellites.
- C.
There are very few microsatellites in eukaryotic genomes so they are easily identified and analyzed.
- D.
Microsatellites are present throughout eukaryotic genomes and are easily amplified using PCR.
Correct Answer
D. Microsatellites are present throughout eukaryotic genomes and are easily amplified using PCR.Explanation
Microsatellites are used more commonly than minisatellites as DNA markers because they are present throughout eukaryotic genomes and can be easily amplified using PCR. This means that they are more readily available and accessible for genetic analysis compared to minisatellites. Additionally, the use of restriction enzymes to type microsatellites is possible, whereas it is not feasible for minisatellites.Rate this question:
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- 37.
Which of the following genetic markers are present in the highest numbers within the human genome?
- A.
RFLPs.
- B.
Minisatellites.
- C.
Microsatellites.
- D.
Single nucleotide polymorphisms.
Correct Answer
C. Microsatellites.Explanation
Microsatellites are the genetic markers that are present in the highest numbers within the human genome. Microsatellites are short repetitive sequences of DNA, typically consisting of 1-6 base pairs, that are scattered throughout the genome. They are highly polymorphic, meaning that they vary in length between individuals due to differences in the number of repeats. This high level of polymorphism makes microsatellites useful for genetic studies, such as determining genetic relatedness, identifying individuals, and studying population genetics. RFLPs, minisatellites, and single nucleotide polymorphisms (SNPs) are also genetic markers, but they are not present in as high numbers as microsatellites in the human genome.Rate this question:
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- 38.
The principle of genetic linkage is:
- A.
The fact that the different alleles for a given gene will be located at the same position in a chromosome.
- B.
The discovery that multiple genes are responsible for some traits (such as eye color in flies).
- C.
The observation that some genes will be inherited together if they are located on the same chromosome.
- D.
The observation that darkly staining regions of chromosomes do not contain genes.
Correct Answer
C. The observation that some genes will be inherited together if they are located on the same chromosome.Explanation
The principle of genetic linkage is the observation that some genes will be inherited together if they are located on the same chromosome. This means that genes that are physically close to each other on a chromosome are more likely to be inherited together during the process of reproduction. This observation led to the understanding that genes are not always inherited independently, but can be linked together based on their physical proximity. This principle forms the basis for understanding patterns of inheritance and genetic mapping.Rate this question:
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- 39.
The difference between mitosis and meiosis is that mitosis is characterized by:
- A.
The production of two diploid cells that are genetically identical to the parental cell.
- B.
The exchange of DNA (crossing-over) between homologous chromosomes.
- C.
The production of two diploid cells that are genetically distinct from the parental cell.
- D.
The production of four haploid celts that are genetically distinct from the parental cell.
Correct Answer
A. The production of two diploid cells that are genetically identical to the parental cell.Explanation
Mitosis is a type of cell division in which a single cell divides into two identical daughter cells. This process results in the production of two diploid cells, meaning they have the same number of chromosomes as the parental cell. Additionally, these cells are genetically identical to the parental cell, as there is no exchange of DNA or crossing-over between homologous chromosomes during mitosis. Therefore, the correct answer is that mitosis is characterized by the production of two diploid cells that are genetically identical to the parental cell.Rate this question:
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- 40.
Which of the following statements correctly describes the recombination frequency between two genes?
- A.
The closer two genes are to each other on a chromosome, the higher the frequency of recombination will be between them.
- B.
The more distant two genes are to each other on a chromosome, the higher the frequency of recombination will be between them.
- C.
If two genes are located on the same chromosome then no recombination events can occur between them.
- D.
If two genes are located on different chromosomes then there will be a high frequency of recombination between them.
Correct Answer
B. The more distant two genes are to each other on a chromosome, the higher the frequency of recombination will be between them.Explanation
The statement that the more distant two genes are to each other on a chromosome, the higher the frequency of recombination will be between them is correct. Recombination is the process by which genetic material is exchanged between homologous chromosomes during meiosis. The farther apart two genes are on a chromosome, the more likely they are to undergo recombination because there is a greater chance for a crossover event to occur between them. This is because the distance between genes on a chromosome determines the likelihood of a crossover event happening during meiosis.Rate this question:
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- 41.
In analyzing a human pedigree to determine how closely two genes are linked, it is best to:
- A.
Conclude that the most common genotypes in the offspring are the parental genotypes
- B.
Conclude that the most common genotypes in the offspring are the recombinants.
- C.
Perform a test cross to determine the linkage between the genes.
- D.
Determine the genotypes of the grandparents.
Correct Answer
D. Determine the genotypes of the grandparents.Explanation
To determine how closely two genes are linked, it is best to determine the genotypes of the grandparents. This is because the genes of interest can be inherited from either parent, and by examining the genotypes of the grandparents, we can determine if the genes are on the same chromosome and therefore linked. If the genes are linked, the most common genotypes in the offspring would be the parental genotypes, rather than the recombinants. Performing a test cross can also help determine linkage, but it is not the best approach in this case.Rate this question:
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- 42.
Which of the following is NOT a factor that limits the accuracy of genetic maps for humans and other complex eukaryotic organisms?
- A.
It is not possible to obtain enough progeny for many eukaryotic organisms.
- B.
Recombination hotspots may interfere with genetic mapping.
- C.
Genetic mapping only uses genes and there are not enough genes to map entire genomes.
- D.
Genes or markers that are tens of thousands of base pairs apart may appear at the same position on a genetic map.
Correct Answer
C. Genetic mapping only uses genes and there are not enough genes to map entire genomes.Explanation
Genetic mapping is a technique used to determine the location of genes on a chromosome. However, it is not limited to just genes but can also include markers or other DNA sequences. The given answer suggests that genetic mapping only uses genes, which is incorrect. In reality, genetic mapping can involve various markers and DNA sequences to map entire genomes. Therefore, this is not a factor that limits the accuracy of genetic maps for humans and other complex eukaryotic organisms.Rate this question:
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- 43.
Metaphase chromosomes were initially used for fluorescent in situ hybridization, but the results were somewhat limiting because:
- A.
Many regions of a chromosome are condensed and cannot hybridize to probes.
- B.
The probes will hybridize preferentially to repeated sequences present on multiple chromosomes.
- C.
The chromosomes are not stable in the condensed state, and the signal diffuses when the chromosomes are relaxed.
- D.
Only low-resolution mapping is possible, as the chromosomes are condensed.
Correct Answer
D. Only low-resolution mapping is possible, as the chromosomes are condensed.Explanation
Metaphase chromosomes are highly condensed and tightly packed, which limits the resolution of mapping. This means that it is difficult to accurately determine the specific locations of genes or DNA sequences on the chromosomes. The other options mentioned in the question, such as regions of the chromosome being unable to hybridize to probes and the signal diffusing when the chromosomes are relaxed, are also limitations of using metaphase chromosomes for fluorescent in situ hybridization. However, the most significant limitation is the low-resolution mapping due to the condensed nature of the chromosomes.Rate this question:
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- 44.
Lnterphase chromosomes are useful for fine-scale mapping by fluorescent in situ hybridization because they:
- A.
Are the least condensed type of chromosome.
- B.
Are easily distinguished from each other by their structures.
- C.
Have transcriptionally active regions of chromatin that are required for this technique.
- D.
Allow for physical mapping of genomes to 1 kb resolution.
Correct Answer
A. Are the least condensed type of chromosome.Explanation
Interphase chromosomes are the least condensed type of chromosome, meaning that they are in a more extended and open form compared to chromosomes during other stages of the cell cycle. This makes them ideal for fine-scale mapping by fluorescent in situ hybridization (FISH) because the less condensed structure allows for easier access to specific DNA sequences. In FISH, fluorescently labeled DNA probes are used to bind to specific target sequences on the chromosomes, and the less condensed structure of interphase chromosomes allows for better visualization and identification of these target sequences.Rate this question:
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- 45.
Sequence tagged sites have which of the following properties?
- A.
They are present only once within a genome and possess an ,RFLP site.
- B.
They are present only once within a genome and their sequence is known.
- C.
Their sequence is known and they must contain repetitive DNA sequences.
- D.
They must contain the sequence of a gene, and no repetitive DNA sequences can be present.
Correct Answer
B. They are present only once within a genome and their sequence is known.Explanation
Sequence tagged sites (STS) are short DNA sequences that are unique and present only once within a genome. These sequences have been identified and their location and sequence are known. STS are useful in genetic mapping and genome sequencing projects as they can be used as landmarks to map other genes or DNA sequences. They are not required to contain repetitive DNA sequences or the sequence of a specific gene. Therefore, the correct answer is that STS are present only once within a genome and their sequence is known.Rate this question:
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- 46.
Which of the following sequences can NOT be used as a sequence tagged site?
- A.
Expressed sequence tags.
- B.
Random genomic sequences.
- C.
Simple sequence length polymorphisms.
- D.
Restriction fragment length polymorphisms.
Correct Answer
D. Restriction fragment length polymorpHisms.Explanation
Restriction fragment length polymorphisms (RFLPs) cannot be used as a sequence tagged site (STS) because RFLPs are variations in the length of DNA fragments caused by genetic mutations or rearrangements, rather than specific DNA sequences. In contrast, expressed sequence tags (ESTs), random genomic sequences, and simple sequence length polymorphisms (SSLPs) are all DNA sequences that can be used as STSs for various genetic analyses.Rate this question:
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- 47.
Radiation hybrid panels provide a useful mechanism for physical mapping because: NOT SURE - STUCK BETWEEN A AND D
- A.
Only a portion of the human genome is present in any given hybrid cell.
- B.
The host hamster cells lack homologous sequences to human genetic markers.
- C.
The host hamster cells are resistant to irradiation.
- D.
A complete physical map of the hamster genome is known.
Correct Answer
A. Only a portion of the human genome is present in any given hybrid cell.Explanation
Radiation hybrid panels are created by fusing human cells with hamster cells, resulting in hybrid cells that contain fragments of the human genome. These hybrid cells can be used to create a physical map of the human genome because each hybrid cell contains only a portion of the human genome. By analyzing the presence or absence of specific genetic markers in the hybrid cells, researchers can determine the order and distance between these markers on the genome. This allows for the construction of a physical map that can be used for various genomic studies.Rate this question:
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- 48.
What would happen if the concentration of dideoxynucleotides was too high in a chain termination sequencing reaction?
- A.
The reactions would yield very long molecules and there would be little sequence data close to the primer.
- B.
The reactions would yield very short molecules.
- C.
The reactions would not proceed as the high concentrations of the dideoxynucleotides would inhibit the DNA polymerase.
- D.
The fluorescence of the sequencing products would be too high and difficult to read.
Correct Answer
B. The reactions would yield very short molecules.Explanation
If the concentration of dideoxynucleotides is too high in a chain termination sequencing reaction, the dideoxynucleotides will be incorporated into the growing DNA chain more frequently, causing premature termination of the chain. This will result in the production of very short DNA molecules, as the sequencing reaction will be terminated at multiple points along the DNA template.Rate this question:
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- 49.
How are the different nucleotides (A, C, G, or T) labeled in a chain termination sequencing reaction?
- A.
The primers for the reactions are labeled with fluorescent dyes.
- B.
The different deoxynucleotides are each labeled with a different fluorescent dye.
- C.
The different dideoxynucleotides are each labeled with a different fluorescent dye.
- D.
The different sequencing products are stained with antibodies that detect the different dideoxynucleotides.
Correct Answer
C. The different dideoxynucleotides are each labeled with a different fluorescent dye.Explanation
In a chain termination sequencing reaction, the different dideoxynucleotides (ddNTPs) are labeled with different fluorescent dyes. This allows for the identification of the nucleotide at each position in the DNA sequence. The ddNTPs are incorporated into the growing DNA chain during the sequencing reaction, causing chain termination. The terminated fragments are then separated by size using capillary electrophoresis, and the fluorescent dye attached to each ddNTP emits a specific wavelength of light, which is detected and used to determine the sequence of the DNA.Rate this question:
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- 50.
Why is it advantageous to clone a DNA fragment prior to chain termination sequencing!
- A.
The chain termination sequencing process requires single-stranded DNA molecules as templates.
- B.
The chain termination sequencing process requires double-stranded DNA molecules as templates.
- C.
The chain termination sequencing process requires a vector to stabilize the template DNA.
- D.
The dideoxynucleotides are only incorporated into cloned DNA fragments.
Correct Answer
A. The chain termination sequencing process requires single-stranded DNA molecules as templates.Explanation
Cloning a DNA fragment prior to chain termination sequencing is advantageous because the chain termination sequencing process requires single-stranded DNA molecules as templates. By cloning the DNA fragment, it can be amplified and converted into multiple copies, allowing for a larger amount of single-stranded DNA templates to be available for sequencing. This increases the sensitivity and efficiency of the sequencing process.Rate this question:
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Mar 21, 2023Quiz Edited by
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Dec 05, 2013Quiz Created by
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