Chapter 16-17 Test - AP Biology
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Refer to the following list of enzymes to answer the following questions. The answers may be used once, more than once, or not at all. A. helicase B. nuclease C. ligase D. DNA polymerase I E. primase separates the DNA strands during replication
This AP Biology quiz assesses knowledge on DNA and protein roles in genetics, covering historical experiments and discoveries from Griffith to Watson and Crick. It evaluates understanding of transformation in bacteria, genetic material complexities, and contributions to DNA structure and function.
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- 2.
Refer to the following list of enzymes to answer the following questions. The answers may be used once, more than once, or not at all. A. helicase B. nuclease C. ligase D. DNA polymerase I E. primase covalently connects segments of DNA
Explanation
DNA ligase is the enzyme responsible for covalently connecting segments of DNA. It plays a crucial role in DNA replication and repair, as well as in recombinant DNA technology, where it is used to join DNA fragments together. The enzyme catalyzes the formation of a phosphodiester bond between the 3' hydroxyl group of one nucleotide and the 5' phosphate group of another nucleotide, thereby linking the DNA segments together.Rate this question:
- 3.
The strands that make up DNA are antiparallel. This means that
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The twisting nature of DNA creates nonparallel strands.
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The 5' to 3' direction of one strand runs counter to the 5' to 3' direction of the other strand.
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Base pairings create unequal spacing between the two DNA strands.
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One strand is positively charged and the other is negatively charged.
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One strand contains only purines and the other contains only pyrimidines.
Correct Answer
A. The 5' to 3' direction of one strand runs counter to the 5' to 3' direction of the other strand.Explanation
The correct answer is that the 5' to 3' direction of one strand runs counter to the 5' to 3' direction of the other strand. This is because DNA strands are antiparallel, meaning they run in opposite directions. One strand has a 5' end and a 3' end, while the other strand has a 3' end and a 5' end. This antiparallel arrangement allows for the complementary base pairing between the strands, where adenine pairs with thymine and guanine pairs with cytosine.Rate this question:
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- 4.
Refer to the following list of enzymes to answer the following questions. The answers may be used once, more than once, or not at all. A. helicase B. nuclease C. ligase D. DNA polymerase I E. primase DNA-cutting enzymes used in the repair of DNA damage
Correct Answer
b
BExplanation
The correct answer is "b, B". The question asks for DNA-cutting enzymes used in the repair of DNA damage. The enzyme "nuclease" is responsible for cutting DNA strands, and it is commonly used in DNA repair processes. Therefore, option b, which includes the enzyme "nuclease", is the correct answer. Option B, which is the same as option b, is also correct as it represents the same enzyme.Rate this question:
- 5.
The Y-shaped structure where the DNA double helix is actively unwound during DNA replication is called the
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Replication fork.
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Replication Y.
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Elongation junction.
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Unwinding point.
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Y junction.
Correct Answer
A. Replication fork.Explanation
The Y-shaped structure where the DNA double helix is actively unwound during DNA replication is called the replication fork. This is the site where the DNA strands separate and new complementary strands are synthesized. The replication fork is formed by the action of enzymes called helicases, which unwind the DNA molecule, and DNA polymerases, which synthesize the new strands. The replication fork moves along the DNA molecule as replication progresses, unwinding and synthesizing new DNA strands in a continuous or discontinuous manner. This process ensures accurate replication of the genetic material.Rate this question:
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- 6.
What kind of chemical bond is found between paired bases of the DNA double helix?
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Hydrogen
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Ionic
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Covalent
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Sulfhydryl
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Phosphate
Correct Answer
A. HydrogenExplanation
The correct answer is hydrogen. Hydrogen bonds are formed between the paired bases of the DNA double helix. These bonds are relatively weak compared to covalent bonds, but they play a crucial role in stabilizing the structure of DNA. The hydrogen bonds occur between specific base pairs, with adenine (A) forming two hydrogen bonds with thymine (T) and guanine (G) forming three hydrogen bonds with cytosine (C). These hydrogen bonds help to hold the two strands of DNA together and allow for the complementary base pairing that is essential for DNA replication and protein synthesis.Rate this question:
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- 7.
Refer to the following list of enzymes to answer the following questions. The answers may be used once, more than once, or not at all. A. helicase B. nuclease C. ligase D. DNA polymerase I E. primase removes the RNA nucleotides from the primer and adds equivalent DNA nucleotides to the 3' end of Okazaki fragments
Correct Answer
d
DExplanation
DNA polymerase I is the enzyme that removes the RNA nucleotides from the primer and adds equivalent DNA nucleotides to the 3' end of Okazaki fragments. This enzyme is responsible for the process of DNA replication, specifically in the synthesis of the lagging strand. It has both exonuclease and polymerase activities, allowing it to remove the RNA primer and replace it with DNA. The correct answer, d,D, refers to DNA polymerase I.Rate this question:
- 8.
Refer to the following list of enzymes to answer the following questions. The answers may be used once, more than once, or not at all. A. helicase B. nuclease C. ligase D. DNA polymerase I E. primase synthesizes short segments of RNA
Correct Answer
e
EExplanation
Enzyme E, also known as primase, synthesizes short segments of RNA. This is an important step in DNA replication, as these RNA segments serve as primers for DNA polymerase to begin synthesizing new DNA strands. Primase adds a short RNA sequence to the DNA template, providing a starting point for DNA polymerase to attach and begin adding nucleotides to form a new DNA strand. Therefore, enzyme E is responsible for the initiation of DNA replication by synthesizing RNA primers.Rate this question:
- 9.
In his transformation experiments, Griffith observed that
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Mutant mice were resistant to bacterial infections.
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Mixing a heat-killed pathogenic strain of bacteria with a living nonpathogenic strain can convert some of the living cells into the pathogenic form.
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Mixing a heat-killed nonpathogenic strain of bacteria with a living pathogenic strain makes the pathogenic strain nonpathogenic.
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Infecting mice with nonpathogenic strains of bacteria makes them resistant to pathogenic strains.
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Mice infected with a pathogenic strain of bacteria can spread the infection to other mice.
Correct Answer
A. Mixing a heat-killed pathogenic strain of bacteria with a living nonpathogenic strain can convert some of the living cells into the pathogenic form.Explanation
Griffith's transformation experiments showed that mixing a heat-killed pathogenic strain of bacteria with a living nonpathogenic strain can convert some of the living cells into the pathogenic form. This implies that there is a transfer of genetic material from the dead pathogenic cells to the living nonpathogenic cells, causing them to become pathogenic. This observation provided evidence for the concept of bacterial transformation, which is the transfer of genetic material between bacteria.Rate this question:
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- 10.
Avery and his colleagues purified various chemicals from pathogenic bacteria and showed that ________ was (were) the transforming agent.
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DNA
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Protein
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Lipids
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Carbohydrates
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Phage
Correct Answer
A. DNAExplanation
Avery and his colleagues purified various chemicals from pathogenic bacteria and demonstrated that DNA was the transforming agent. This suggests that DNA is responsible for transferring genetic information and causing hereditary changes in bacteria. This finding was significant in understanding the role of DNA in genetics and further supported the idea that DNA is the genetic material.Rate this question:
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- 11.
A space probe returns with a culture of a microorganism found on a distant planet. Analysis shows that it is a carbon-based life-form that has DNA. You grow the cells in 15N medium for several generations and then transfer them to 14N medium. Which pattern in Figure 16.1 would you expect if the DNA was replicated in a conservative manner?
Correct Answer
b
BExplanation
If the DNA was replicated in a conservative manner, it means that after each replication, one of the daughter DNA molecules would consist entirely of newly synthesized strands, while the other would consist entirely of the original parental strands. This would result in a pattern where one DNA molecule would have all 15N nucleotides (b) and the other DNA molecule would have all 14N nucleotides (B).Rate this question:
- 12.
What is the role of DNA ligase in the elongation of the lagging strand during DNA replication?
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Synthesize RNA nucleotides to make a primer
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Catalyze the lengthening of telomeres
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Join Okazaki fragments together
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Unwind the parental double helix
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Stabilize the unwound parental DNA
Correct Answer
A. Join Okazaki fragments togetherExplanation
During DNA replication, the lagging strand is synthesized in short fragments called Okazaki fragments. DNA ligase plays a crucial role in the elongation of the lagging strand by joining these Okazaki fragments together. It catalyzes the formation of phosphodiester bonds between the adjacent nucleotides, effectively sealing the gaps between the fragments and creating a continuous strand. This process ensures that both the leading and lagging strands are fully synthesized and complete.Rate this question:
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- 13.
What is the function of DNA polymerase?
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To unwind the DNA helix during replication
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To seal together the broken ends of DNA strands
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To add nucleotides to the end of a growing DNA strand
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To degrade damaged DNA molecules
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To rejoin the two DNA strands (one new and one old) after replication
Correct Answer
A. To add nucleotides to the end of a growing DNA strandExplanation
DNA polymerase is an enzyme responsible for adding nucleotides to the end of a growing DNA strand. During DNA replication, DNA polymerase attaches to the template strand and adds complementary nucleotides to the growing daughter strand, following the base pairing rules (A with T, G with C). This process ensures that the new DNA strand is an exact copy of the original template strand. DNA polymerase plays a crucial role in DNA replication, as it helps to maintain the integrity and accuracy of the genetic information.Rate this question:
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- 14.
Which of the following represents a similarity between RNA and DNA?
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Both are double-stranded.
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The presence of uracil
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The presence of an OH group on the 2' carbon of the sugar
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Nucleotides consisting of a phosphate, sugar, and nitrogenous base
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Both are found exclusively in the nucleus.
Correct Answer
A. Nucleotides consisting of a phosphate, sugar, and nitrogenous baseExplanation
Both RNA and DNA are made up of nucleotides consisting of a phosphate group, a sugar molecule, and a nitrogenous base. This similarity is fundamental to their structure and function as genetic material. The presence of uracil in RNA and the presence of an OH group on the 2' carbon of the sugar in RNA are specific characteristics of RNA and not shared by DNA. Additionally, both RNA and DNA can be found not only in the nucleus but also in other parts of the cell.Rate this question:
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- 15.
A codon
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Consists of two nucleotides.
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May code for the same amino acid as another codon
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Consists of discrete amino acid regions.
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Catalyzes RNA synthesis.
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Is found in all eukaryotes, but not in prokaryotes.
Correct Answer
A. May code for the same amino acid as another codonExplanation
Codons are sequences of three nucleotides that code for specific amino acids during protein synthesis. There are multiple codons that can code for the same amino acid, which is known as degeneracy in the genetic code. This allows for redundancy and flexibility in the genetic code, reducing the impact of mutations or errors during DNA replication. Therefore, it is possible for one codon to code for the same amino acid as another codon.Rate this question:
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- 16.
The DNA double helix has a uniform diameter because ________, which have two rings, always pair with ________, which have one ring.
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Purines; pyrimidines
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Pyrimidines; purines
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Deoxyribose sugars; ribose sugars
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Ribose sugars; deoxyribose sugars
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Nucleotides; nucleoside triphosphates
Correct Answer
A. Purines; pyrimidinesExplanation
The DNA double helix has a uniform diameter because purines, which have two rings, always pair with pyrimidines, which have one ring. This pairing allows the two strands of DNA to fit together perfectly, maintaining a consistent diameter throughout the molecule.Rate this question:
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- 17.
In the late 1950s, Meselson and Stahl grew bacteria in a medium containing "heavy" nitrogen (15N) and then transferred them to a medium containing 14N. Which of the above results would be expected after one DNA replication in the presence of 14N?
Correct Answer
D
dExplanation
After one DNA replication in the presence of 14N, it is expected that the DNA strands will consist of one heavy (15N) strand and one light (14N) strand. This is because during DNA replication, the original DNA molecule serves as a template for the synthesis of a new complementary strand. Therefore, each original heavy strand will be replicated to produce one heavy strand and one newly synthesized light strand. Hence, the expected result after one DNA replication in the presence of 14N is a mixture of DNA molecules with one heavy strand and one light strand.Rate this question:
- 18.
You briefly expose bacteria undergoing DNA replication to radioactively labeled nucleotides. When you centrifuge the DNA isolated from the bacteria, the DNA separates into two classes. One class of labeled DNA includes very large molecules (thousands or even millions of nucleotides long), and the other includes short stretches of DNA (several hundred to a few thousand nucleotides in length). These two classes of DNA probably represent
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Leading strands and Okazaki fragments.
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Lagging strands and Okazaki fragments
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Okazaki fragments and RNA primers.
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Leading strands and RNA primers.
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RNA primers and mitochondrial DNA.
Correct Answer
A. Leading strands and Okazaki fragments.Explanation
The presence of very large molecules in one class of labeled DNA suggests that these molecules are the leading strands, which are synthesized continuously during DNA replication. The presence of short stretches of DNA in the other class indicates that these are Okazaki fragments, which are synthesized discontinuously on the lagging strand. This is because the lagging strand is synthesized in short fragments that are later joined together. Therefore, the two classes of DNA likely represent leading strands and Okazaki fragments.Rate this question:
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- 19.
Once transcribed, eukaryotic mRNA typically undergoes substantial alteration that includes
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Excision of introns.
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Fusion into circular forms known as plasmids.
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Linkage to histone molecules.
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Union with ribosomes.
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Fusion with other newly transcribed mRNA.
Correct Answer
A. Excision of introns.Explanation
Eukaryotic mRNA undergoes substantial alteration after transcription, and one of these alterations is the excision of introns. Introns are non-coding regions within the mRNA that do not contain information for protein synthesis. They are removed through a process called splicing, which results in the production of mature mRNA that only contains the coding regions called exons. This allows for the production of functional proteins. The other options provided in the question, such as fusion into circular forms, linkage to histone molecules, union with ribosomes, or fusion with other mRNA, are not accurate explanations for the alterations that occur in eukaryotic mRNA.Rate this question:
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- 20.
Cytosine makes up 38% of the nucleotides in a sample of DNA from an organism. Approximately, what percentage of the nucleotides in this sample will be thymine?
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12
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24
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31
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38
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It cannot be determined from the information provided.
Correct Answer
A. 24Explanation
In DNA, cytosine (C) always pairs with guanine (G), and thymine (T) always pairs with adenine (A). The percentage of cytosine (C) in DNA will be the same as the percentage of guanine (G). Therefore, the percentage of thymine (T) will be the same as the percentage of adenine (A).
If cytosine makes up 38% of the nucleotides, then guanine also makes up 38% of the nucleotides.
The total percentage of cytosine (C) and guanine (G) combined is 38% + 38% = 76%.
Since the total percentage of nucleotides is 100%, and cytosine (C) and guanine (G) together make up 76%, the remaining percentage for thymine (T) and adenine (A) combined will be 100% - 76% = 24%.
Because adenine (A) pairs with thymine (T), the percentage of thymine (T) will be the same as the percentage of adenine (A), which is 24%.
Therefore, approximately 24% of the nucleotides in this sample will be thymine (T).Rate this question:
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- 21.
Which of the following help to hold the DNA strands apart while they are being replicated?
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Primase
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Ligase
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DNA polymerase
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Single-strand binding proteins
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Exonuclease
Correct Answer
A. Single-strand binding proteinsExplanation
Single-strand binding proteins help to hold the DNA strands apart while they are being replicated. These proteins bind to the single-stranded DNA and prevent the strands from reannealing or forming secondary structures, allowing the replication machinery to access the DNA template and synthesize a new complementary strand. Primase is responsible for synthesizing RNA primers, ligase joins the Okazaki fragments, DNA polymerase synthesizes the new DNA strands, and exonuclease removes nucleotides from the ends of DNA strands. However, none of these directly hold the DNA strands apart during replication.Rate this question:
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- 22.
The nitrogenous base adenine is found in all members of which group?
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Proteins, triglycerides, and testosterone
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Proteins, ATP, and DNA
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ATP, RNA, and DNA
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Alpha glucose, ATP, and DNA
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Proteins, carbohydrates, and ATP
Correct Answer
A. ATP, RNA, and DNAExplanation
Adenine is a nitrogenous base that is found in ATP, RNA, and DNA. Adenine is one of the four bases that make up the genetic code in DNA and RNA. It is also a component of ATP, which is the primary energy source for cellular processes. Therefore, the correct answer is ATP, RNA, and DNA.Rate this question:
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- 23.
If the triplet CCC codes for the amino acid proline in bacteria, then in plants CCC should code for
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Leucine.
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Valine.
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Cystine.
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Phenylalanine.
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Proline.
Correct Answer
A. Proline.Explanation
In bacteria, the triplet CCC codes for the amino acid proline. Since the question states that this is the case in bacteria, it can be inferred that in plants, CCC should also code for proline. Therefore, the correct answer is proline.Rate this question:
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- 24.
What is an anticodon part of?
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DNA
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TRNA
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MRNA
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A ribosome
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An activating enzyme
Correct Answer
A. TRNAExplanation
An anticodon is a part of tRNA (transfer RNA). tRNA is responsible for carrying specific amino acids to the ribosome during protein synthesis. The anticodon is a sequence of three nucleotides on the tRNA molecule that is complementary to a specific codon on the mRNA (messenger RNA). This allows the tRNA to recognize and bind to the correct codon on the mRNA, ensuring that the correct amino acid is added to the growing protein chain. Therefore, the anticodon plays a crucial role in the accurate translation of the genetic code.Rate this question:
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- 25.
Which enzyme catalyzes the elongation of a DNA strand in the 5' ïƒ 3' direction?
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Primase
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DNA ligase
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DNA polymerase
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Topoisomerase
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Helicase
Correct Answer
A. DNA polymeraseExplanation
DNA polymerase is the enzyme responsible for catalyzing the elongation of a DNA strand in the 5' to 3' direction. This enzyme adds nucleotides to the growing DNA strand, using the template strand as a guide. It also proofreads the newly synthesized DNA strand to ensure accuracy. Primase is involved in synthesizing RNA primers, DNA ligase joins Okazaki fragments during DNA replication, topoisomerase relieves tension in the DNA molecule, and helicase unwinds the DNA double helix.Rate this question:
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- 26.
Which of these mechanisms ensures that the DNA sequence in the genome remains accurate?
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Proofreading during DNA replication
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Mismatch repair
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Excision repair
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Complementary base pairing during DNA replication
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All of the above
Correct Answer
A. All of the aboveExplanation
All of the mechanisms mentioned in the options ensure that the DNA sequence in the genome remains accurate. Proofreading during DNA replication involves the correction of errors made during the copying of DNA. Mismatch repair is a process that corrects errors that occur after DNA replication. Excision repair is a mechanism that repairs DNA damage caused by environmental factors. Complementary base pairing during DNA replication ensures that the new DNA strand is accurately synthesized based on the template strand. Therefore, all of these mechanisms work together to maintain the accuracy of the DNA sequence in the genome.Rate this question:
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- 27.
For a science fair project, two students decided to repeat the Hershey and Chase experiment, with modifications. They decided to label the nitrogen of the DNA, rather than the phosphate. They reasoned that each nucleotide has only one phosphate and two to five nitrogens. Thus, labeling the nitrogens would provide a stronger signal than labeling the phosphates. Why won't this experiment work?
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There is no radioactive isotope of nitrogen.
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Radioactive nitrogen has a half-life of 100,000 years, and the material would be too dangerous for too long.
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Meselson and Stahl already did this experiment.
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Although there are more nitrogens in a nucleotide, labeled phosphates actually have 16 extra neutrons; therefore, they are more radioactive.
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Amino acids (and thus proteins) also have nitrogen atoms; thus, the radioactivity would not distinguish between DNA and proteins.
Correct Answer
A. Amino acids (and thus proteins) also have nitrogen atoms; thus, the radioactivity would not distinguish between DNA and proteins.Explanation
Labeling the nitrogens of the DNA instead of the phosphates would not work because amino acids and proteins also have nitrogen atoms. Therefore, the radioactivity from the labeled nitrogens would not be able to distinguish between DNA and proteins.Rate this question:
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- 28.
It became apparent to Watson and Crick after completion of their model that the DNA molecule could carry a vast amount of hereditary information in its
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Sequence of bases.
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Phosphate-sugar backbones.
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Complementary pairing of bases.
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Side groups of nitrogenous bases.
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Different five-carbon sugars.
Correct Answer
A. Sequence of bases.Explanation
After completing their model of the DNA molecule, Watson and Crick realized that the sequence of bases in the DNA molecule could carry a vast amount of hereditary information. The sequence of bases, which includes adenine (A), thymine (T), cytosine (C), and guanine (G), determines the genetic code and is responsible for encoding the instructions for building and maintaining an organism. This discovery was crucial in understanding how DNA carries and transmits genetic information.Rate this question:
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- 29.
What amino acid sequence will be generated, based on the following mRNA codon sequence? 5'AUG-UCU-UCG-UUA-UCC-UUG
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Met-arg-glu-arg-glu-arg
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Met-glu-arg-arg-gln-leu
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Met-ser-leu-ser-leu-ser
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Met-ser-ser-leu-ser-leu
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Met-leu-phe-arg-glu-glu
Correct Answer
A. Met-ser-ser-leu-ser-leuExplanation
The mRNA codon sequence 5'AUG-UCU-UCG-UUA-UCC-UUG' translates to the amino acid sequence 'met-ser-ser-leu-ser-leu'. This is because the codon 'AUG' codes for the start codon methionine (met), 'UCU' and 'UCC' both code for serine (ser), 'UCG' codes for serine (ser), 'UUA' and 'UUG' both code for leucine (leu). Therefore, the correct answer is 'met-ser-ser-leu-ser-leu'.Rate this question:
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- 30.
Which of the following helps to stabilize mRNA by inhibiting its degradation?
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TATA box
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Spliceosomes
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5' cap
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Poly-A tail
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Both C and D
Correct Answer
A. Both C and DExplanation
The 5' cap and poly-A tail both help to stabilize mRNA by inhibiting its degradation. The 5' cap is added to the 5' end of the mRNA molecule and protects it from exonucleases, which degrade RNA from the ends. The poly-A tail is added to the 3' end of the mRNA molecule and helps to protect it from degradation by exonucleases as well. Therefore, both the 5' cap and poly-A tail play important roles in stabilizing mRNA and preventing its degradation.Rate this question:
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- 31.
All of the following can be determined directly from X-ray diffraction photographs of crystallized DNA except the
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Diameter of the helix.
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Helical shape of DNA.
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Sequence of nucleotides.
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Spacing of the nitrogenous bases along the helix.
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Number of strands in a helix.
Correct Answer
A. Sequence of nucleotides.Explanation
X-ray diffraction photographs of crystallized DNA provide information about the helical shape of DNA, the spacing of the nitrogenous bases along the helix, and the number of strands in a helix. However, determining the sequence of nucleotides requires other techniques such as DNA sequencing methods. Therefore, the sequence of nucleotides cannot be directly determined from X-ray diffraction photographs of crystallized DNA.Rate this question:
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- 32.
What kind of molecule or substance is the primer that is used to initiate the synthesis of a new DNA strand?
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RNA
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DNA
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Protein
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Phosphate
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Sulfur
Correct Answer
A. RNAExplanation
The primer used to initiate the synthesis of a new DNA strand is RNA. RNA primers are short sequences of RNA that are synthesized by an enzyme called primase. These primers provide a starting point for DNA polymerase to begin synthesizing the complementary DNA strand during DNA replication. RNA primers are later removed and replaced with DNA by another enzyme called DNA polymerase, resulting in a complete double-stranded DNA molecule. Therefore, RNA is the correct answer as it serves as the initial template for DNA synthesis.Rate this question:
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- 33.
What is the function of topoisomerase?
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Relieving strain in the DNA ahead of the replication fork
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Elongation of new DNA at a replication fork by addition of nucleotides to the existing chain
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The addition of methyl groups to bases of DNA
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Unwinding of the double helix
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Stabilizing single-stranded DNA at the replication fork
Correct Answer
A. Relieving strain in the DNA ahead of the replication forkExplanation
Topoisomerase is responsible for relieving strain in the DNA ahead of the replication fork. During DNA replication, the double helix structure of DNA becomes tightly wound and twisted, causing strain and tension. Topoisomerase enzymes work by breaking and rejoining the DNA strands, allowing the DNA to unwind and relieve the strain. This process is crucial for the smooth progression of DNA replication and prevents DNA damage or breakage.Rate this question:
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- 34.
Which of the following is analogous to telomeres?
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The pull tab on a soft drink can
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The two ends of a shoelace
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The central spindle that a CD fits around while in the case
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The mechanism of a zipper that allows the separated parts to be joined
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The correct letters used to replace errors in a document after they have been deleted in a word processor
Correct Answer
A. The two ends of a shoelaceExplanation
Telomeres are protective caps at the ends of chromosomes that prevent them from deteriorating or fusing with neighboring chromosomes. Similarly, the two ends of a shoelace have protective caps called aglets that prevent the lace from fraying and make it easier to thread through the eyelets of a shoe. Both telomeres and shoelace ends serve a similar function of protecting and maintaining the integrity of their respective structures.Rate this question:
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- 35.
Which of the following statements about telomeres is correct?
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They contain multiple copies of a short RNA sequence.
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They are present at the ends of eukaryotic chromosomes.
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They can be extended by an enzyme called telomerase.
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Both A and B
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Both B and C
Correct Answer
A. Both B and CExplanation
Telomeres are repetitive DNA sequences found at the ends of eukaryotic chromosomes. They protect the chromosomes from degradation and fusion with other chromosomes. Telomeres can be extended by an enzyme called telomerase, which adds additional repetitive sequences to the ends of chromosomes. Therefore, the correct statement is that telomeres are present at the ends of eukaryotic chromosomes and can be extended by telomerase.Rate this question:
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- 36.
What is the sequence of a peptide based on the mRNA sequence 5' UUUUCUUAUUGUCUU 3' ?
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Leu-cys-tyr-ser-phe
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Cyc-phe-tyr-cys-leu
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Phe-leu-ile-met-val
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Leu-pro-asp-lys-gly
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Phe-ser-tyr-cys-leu
Correct Answer
A. Phe-ser-tyr-cys-leuExplanation
The mRNA sequence 5' UUUUCUUAUUGUCUU 3' can be translated into the peptide sequence phe-ser-tyr-cys-leu. This is determined by using the genetic code, where each three nucleotides (codon) in the mRNA sequence corresponds to a specific amino acid. In this case, UUU codes for phenylalanine (phe), UCU codes for serine (ser), UUA codes for leucine (leu), UUG codes for leucine (leu), UGU codes for cysteine (cys), and UUC codes for leucine (leu). Therefore, the correct peptide sequence is phe-ser-tyr-cys-leu.Rate this question:
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- 37.
For a couple of decades, biologists knew the nucleus contained DNA and proteins. The prevailing opinion was that the genetic material was proteins, and not DNA. The reason for this belief was that proteins are more complex than DNA. This is because
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Proteins have a greater variety of three-dimensional forms than does DNA.
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Proteins have two different levels of structural organization; DNA has four.
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Proteins are made of 20 amino acids and DNA is made of four nucleotides.
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Only A and C are correct.
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A, B, and C are correct.
Correct Answer
A. Only A and C are correct.Explanation
The prevailing opinion that proteins were the genetic material instead of DNA was based on the belief that proteins are more complex than DNA. This is because proteins have a greater variety of three-dimensional forms and are made up of 20 different amino acids, while DNA is made up of only four nucleotides. Therefore, options A and C are correct in explaining why proteins were initially thought to be the genetic material.Rate this question:
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- 38.
All of the following are found in prokaryotic mRNA except
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The AUG codon.
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The UGA codon.
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Introns.
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Uracil.
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Cytosine.
Correct Answer
A. Introns.Explanation
Prokaryotic mRNA does not contain introns, which are non-coding sections of DNA that are removed during the process of mRNA maturation in eukaryotic cells. Prokaryotes lack the machinery to remove introns, so their mRNA is typically composed solely of coding sequences. The AUG codon is the start codon that signals the beginning of protein synthesis, while the UGA codon is a stop codon that signals the end of protein synthesis. Uracil and cytosine are both nucleotide bases found in RNA.Rate this question:
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- 39.
What determines the nucleotide sequence of the newly synthesized strand during DNA replication?
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The particular DNA polymerase catalyzing the reaction
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The relative amounts of the four nucleoside triphosphates in the cell
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The nucleotide sequence of the template strand
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The primase used in the reaction
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Both A and D
Correct Answer
A. The nucleotide sequence of the template strandExplanation
The nucleotide sequence of the template strand determines the nucleotide sequence of the newly synthesized strand during DNA replication. The template strand serves as a guide for the DNA polymerase to match complementary nucleotides and create a complementary strand. The DNA polymerase adds nucleotides in a sequence that corresponds to the template strand, resulting in a newly synthesized strand with the same nucleotide sequence as the template strand. The other options, such as the particular DNA polymerase, the relative amounts of nucleoside triphosphates, and the primase used, may have roles in DNA replication but do not directly determine the nucleotide sequence of the newly synthesized strand.Rate this question:
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- 40.
Individuals with the disorder xeroderma pigmentosum are hypersensitive to sunlight because their cells have an impaired ability to
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Replicate DNA.
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Undergo mitosis.
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Exchange DNA with other cells.
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Repair thymine dimers.
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Recombine homologous chromosomes during meiosis.
Correct Answer
A. Repair thymine dimers.Explanation
Individuals with xeroderma pigmentosum have a genetic disorder that impairs their ability to repair thymine dimers. Thymine dimers are a type of DNA damage that is caused by exposure to ultraviolet (UV) radiation from sunlight. When these dimers are not repaired, they can lead to mutations in the DNA, which can increase the risk of skin cancer. Therefore, individuals with xeroderma pigmentosum are hypersensitive to sunlight because their cells cannot effectively repair thymine dimers, making them more susceptible to DNA damage and the harmful effects of UV radiation.Rate this question:
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- 41.
Which of the following statements does not apply to the Watson and Crick model of DNA?
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The two strands of the DNA form a double helix.
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The distance between the strands of the helix is uniform.
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The framework of the helix consists of sugar-phosphate units of the nucleotides.
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The two strands of the helix are held together by covalent bonds.
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The purines form hydrogen bonds with pyrimidines.
Correct Answer
A. The two strands of the helix are held together by covalent bonds.Explanation
The Watson and Crick model of DNA states that the two strands of the DNA form a double helix, the distance between the strands of the helix is uniform, the framework of the helix consists of sugar-phosphate units of the nucleotides, and the purines form hydrogen bonds with pyrimidines. However, the model does not propose that the two strands of the helix are held together by covalent bonds. Instead, they are held together by hydrogen bonds between the nitrogenous bases.Rate this question:
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- 42.
Garrod hypothesized that "inborn errors of metabolism" such as alkaptonuria occur because
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Genes dictate the production of specific enzymes, and affected individuals have genetic defects that cause them to lack certain enzymes.
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Enzymes are made of DNA, and affected individuals lack DNA polymerase.
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Many metabolic enzymes use DNA as a cofactor, and affected individuals have mutations that prevent their enzymes from interacting efficiently with DNA.
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Certain metabolic reactions are carried out by ribozymes, and affected individuals lack key splicing factors.
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Metabolic enzymes require vitamin cofactors, and affected individuals have significant nutritional deficiencies.
Correct Answer
A. Genes dictate the production of specific enzymes, and affected individuals have genetic defects that cause them to lack certain enzymes.Explanation
Garrod hypothesized that "inborn errors of metabolism" occur because genes dictate the production of specific enzymes. This means that affected individuals have genetic defects that cause them to lack certain enzymes. This explanation aligns with the concept that genetic mutations can lead to enzyme deficiencies, resulting in metabolic disorders such as alkaptonuria.Rate this question:
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- 43.
Accuracy in the translation of mRNA into the primary structure of a protein depends on specificity in the
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Binding of ribosomes to mRNA.
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Shape of the A and P sites of ribosomes.
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Bonding of the anticodon to the codon.
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Attachment of amino acids to tRNAs.
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Both C and D
Correct Answer
A. Both C and DExplanation
The accuracy in the translation of mRNA into the primary structure of a protein depends on both the bonding of the anticodon to the codon and the attachment of amino acids to tRNAs. The bonding of the anticodon to the codon ensures that the correct amino acid is added to the growing protein chain. The attachment of amino acids to tRNAs ensures that the correct amino acid is available to be added to the growing protein chain. Both of these processes are crucial for the accurate translation of mRNA into a functional protein.Rate this question:
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- 44.
What are ribosomes composed of?
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RRNA only
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Proteins only
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Both rRNA and protein
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MRNA, rRNA, and protein
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MRNA, tRNA, rRNA, and protein
Correct Answer
A. Both rRNA and proteinExplanation
Ribosomes are composed of both ribosomal RNA (rRNA) and proteins. Ribosomes are cellular structures responsible for protein synthesis. They consist of two subunits, a small subunit and a large subunit, which come together to form a functional ribosome. The small subunit contains rRNA and proteins, while the large subunit contains rRNA and proteins as well. The rRNA provides the structural framework for the ribosome, while the proteins play a role in catalyzing the formation of peptide bonds between amino acids during protein synthesis. Therefore, ribosomes are composed of both rRNA and proteins.Rate this question:
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- 45.
The following scientists made significant contributions to our understanding of the structure and function of DNA. Place the scientists' names in the correct chronological order, starting with the first scientist(s) to make a contribution. I. Avery, McCarty, and MacLeod II. Griffith III. Hershey and Chase IV. Meselson and Stahl V. Watson and Crick
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V, IV, II, I, III
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II, I, III, V, IV
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I, II, III, V, IV
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I, II, V, IV, III
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II, III, IV, V, I
Correct Answer
A. II, I, III, V, IVExplanation
The correct answer is II, I, III, V, IV. This represents the chronological order of the scientists' contributions to our understanding of the structure and function of DNA. Griffith was the first to make a contribution, followed by Avery, McCarty, and MacLeod. Hershey and Chase made their contribution next, followed by Watson and Crick. Meselson and Stahl made their contribution last.Rate this question:
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- 46.
Replicating the lagging strand of DNA-that is, adding bases in the 3' ïƒ 5' direction-utilizes which of the following?
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DNA ligase
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RNA primers
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Okazaki fragments
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A and B only
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A, B, and C
Correct Answer
A. A, B, and CExplanation
The replication of the lagging strand of DNA, which involves adding bases in the 3' -> 5' direction, utilizes DNA ligase, RNA primers, and Okazaki fragments. DNA ligase is responsible for joining the Okazaki fragments together, while RNA primers are necessary for initiating DNA synthesis. Okazaki fragments are short DNA fragments that are synthesized on the lagging strand during DNA replication. Therefore, all three options (A, B, and C) are required for the replication of the lagging strand of DNA.Rate this question:
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- 47.
Introns are significant to biological evolution because
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Their presence allows exons to be moved around more easily, creating proteins with new combinations of functional domains.
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They protect the mRNA from degeneration.
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They are translated into essential amino acids.
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They maintain the genetic code by preventing incorrect DNA base pairings.
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They correct enzymatic alterations of DNA bases.
Correct Answer
A. Their presence allows exons to be moved around more easily, creating proteins with new combinations of functional domains.Explanation
Introns are non-coding regions of DNA that are transcribed into RNA but are not translated into proteins. They are significant to biological evolution because their presence allows exons (coding regions) to be rearranged and combined in different ways during the process of alternative splicing. This rearrangement can lead to the creation of proteins with new combinations of functional domains, allowing for increased diversity and adaptation in organisms.Rate this question:
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- 48.
A mutation in which of the following parts of a gene is likely to be most damaging to a cell?
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Intron
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Exon
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5' UTR
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3' UTR
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All would be equally damaging.
Correct Answer
A. ExonExplanation
A mutation in the exon of a gene is likely to be most damaging to a cell because exons are the coding regions of a gene that contain the instructions for producing a functional protein. Mutations in exons can lead to changes in the protein sequence, altering its structure or function. In contrast, introns are non-coding regions that are removed during RNA processing, while the 5' UTR and 3' UTR are untranslated regions that play regulatory roles but do not directly affect the protein sequence. Therefore, mutations in these regions are less likely to have a significant impact on the cell.Rate this question:
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- 49.
All of the following are directly involved in translation except
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MRNA.
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TRNA.
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Ribosomes.
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DNA.
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Aminoacyl-tRNA synthetase enzymes.
Correct Answer
A. DNA.Explanation
Translation is the process by which the genetic information in mRNA is used to synthesize proteins. mRNA carries the genetic code from DNA to the ribosomes, where the actual translation takes place. Ribosomes are the cellular structures responsible for protein synthesis, and tRNA molecules bring the amino acids to the ribosomes based on the codons on the mRNA. Aminoacyl-tRNA synthetase enzymes are responsible for attaching the correct amino acid to the corresponding tRNA molecule. However, DNA is not directly involved in translation. It serves as the template for mRNA synthesis during transcription.Rate this question:
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.JPG "A space probe returns with a culture of a microorganism found on a distant planet. Analysis shows that it is a carbon-based life-form that has DNA. You grow the cells in 15N medium for several generations and then transfer them to 14N medium. Which pattern in Figure 16.1 would you expect if the DNA was replicated in a conservative manner?")
.JPG "In the late 1950s, Meselson and Stahl grew bacteria in a medium containing \"heavy\" nitrogen (15N) and then transferred them to a medium containing 14N. Which of the above results would be expected after one DNA replication in the presence of 14N?")
.JPG "What amino acid sequence will be generated, based on the following mRNA codon sequence? 5'AUG-UCU-UCG-UUA-UCC-UUG")
.JPG "What is the sequence of a peptide based on the mRNA sequence 5' UUUUCUUAUUGUCUU 3' ?")
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Apr 03, 2024Quiz Edited by
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Feb 26, 2013Quiz Created by
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