Genetics Final Exam

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  • 1/100 Questions

    In a pea plant that is heterozygous for seed color, what proportion of gametes will carry the recessive allele?

    • 1/4
    • 1/2
    • 1/3
    • All of the gametes
    • None of the gametes
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About This Quiz

You might wonder why it’s so important to analyze the small, seemingly insignificant details of a person’s genetic make-up. But what you might not realize is that some things about ourselves can’t be seen by the naked eye – like a person’s chances of developing a terminal illness as a result of it being passed down from parent to offspring. See moreWhat can you tell us about genetics?

Genetics Final Exam - Quiz

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  • 2. 

    A(n)                  gene masks the expression of a different, nonallelic gene.

    • Imcompletely dominant

    • Epistatic

    • Recessive

    • Codominant

    • Hemizygous

    Correct Answer
    A. Epistatic
    Explanation
    Epistasis refers to the interaction between two or more genes where the expression of one gene masks or modifies the expression of another gene. In this case, the gene in question is masking the expression of a different, nonallelic gene, indicating an epistatic relationship. This means that the gene is exerting control over the expression of the other gene, overriding its normal function or expression pattern. Therefore, the correct answer is epistatic.

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  • 3. 

    The genetic material of eukaryotic cells is duplicated during which stage of cell cycle?

    • M

    • S

    • G1

    • G2

    • None of the above

    Correct Answer
    A. S
    Explanation
    During the S phase of the cell cycle, the genetic material of eukaryotic cells is duplicated. This phase is also known as the synthesis phase, where DNA replication occurs. The cell prepares itself for division by creating an identical copy of its DNA. This ensures that each daughter cell will have a complete set of genetic material. The S phase is followed by the G2 phase, where the cell prepares for cell division, and then the M phase, where the actual cell division takes place. Therefore, the correct answer is S.

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  • 4. 

    The appearance of traits expressed by genes in association with environmental influence is referred to as the organism's 

    • Genome

    • Karyotype

    • Phenotype

    • Genotype

    • Prototype

    Correct Answer
    A. Phenotype
    Explanation
    The appearance of traits expressed by genes in association with environmental influence is referred to as the organism's phenotype. The phenotype is the observable characteristics of an organism, such as its physical appearance, behavior, and other traits. It is determined by the interaction between an organism's genotype (its genetic makeup) and its environment. The phenotype can be influenced by various factors, including genetic mutations, gene expression, and environmental factors such as diet, temperature, and exposure to toxins.

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  • 5. 

    A man with blood type A and a woman with blood type B could never produce a child with blood type

    • A

    • B

    • AB

    • O

    • None of the above

    Correct Answer
    A. None of the above
    Explanation
    When a man with blood type A and a woman with blood type B have a child, there is a possibility that the child could inherit either blood type A or B from their parents. This is because blood type A is determined by having either two A alleles or one A and one O allele, while blood type B is determined by having either two B alleles or one B and one O allele. Therefore, it is possible for the child to have blood type A, B, or AB, making "none of the above" the correct answer.

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  • 6. 

    In an interacting gene pair, the gene whose expression is masked by another gene is the         gene.

    • Epistatic

    • Homostatic

    • Hypostatic

    • Hyperstatic

    • Limnostatic

    Correct Answer
    A. Hypostatic
    Explanation
    In an interacting gene pair, the gene whose expression is masked by another gene is called the hypostatic gene. This means that the hypostatic gene's effects are suppressed or overridden by the other gene in the pair.

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  • 7. 

    In human males, genes on the X chromosome are

    • Homozygous

    • Heterozygous

    • Homogametic

    • Heterogametic

    • Hemizygous

    Correct Answer
    A. Hemizygous
    Explanation
    In human males, genes on the X chromosome are hemizygous. This is because males have one X chromosome and one Y chromosome, while females have two X chromosomes. Therefore, males have only one copy of genes on the X chromosome, making them hemizygous.

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  • 8. 

    A Barr body is an

    • Active X chromosome

    • Inactive Y chromosome

    • Inactive X chromosome

    • Active Y chromosome

    Correct Answer
    A. Inactive X chromosome
    Explanation
    A Barr body is a condensed, inactive X chromosome that is typically found in the nuclei of female cells. In females, one of the two X chromosomes is randomly inactivated during early development to ensure dosage compensation between males and females. This inactivated X chromosome becomes highly compacted and forms a dense structure known as a Barr body. The presence of a Barr body can be used to identify the sex of an individual's cells, as males typically have one X chromosome and females have two, with one being inactivated.

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  • 9. 

    The basic structure of nucleotide includes the following components

    • Amino acids

    • Base, sugar, phosphate

    • Tryptophan and leucine

    • MRNA, rRNA, tRNA

    • Phosphorus and sulfur

    Correct Answer
    A. Base, sugar, phosphate
    Explanation
    The correct answer is base, sugar, phosphate. Nucleotides are the building blocks of DNA and RNA, and they consist of three main components: a nitrogenous base (adenine, guanine, cytosine, or thymine/uracil), a sugar molecule (deoxyribose in DNA or ribose in RNA), and a phosphate group. These components combine to form a nucleotide, which then join together to form the DNA and RNA strands. The base pairs in DNA (adenine with thymine and guanine with cytosine) and the sequence of nucleotides in RNA determine the genetic information encoded in these molecules.

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  • 10. 

    Bacterial Recombination requires?

    • F pilus

    • Plasmid

    • Physical contact

    • All of the above

    • None of the above

    Correct Answer
    A. All of the above
    Explanation
    Bacterial recombination requires all of the above options because F pilus is necessary for the transfer of genetic material between bacteria, plasmids are the carriers of the genetic material being transferred, and physical contact between bacteria is required for the transfer to occur. Therefore, all three options are essential for bacterial recombination to take place.

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  • 11. 

    The two major components of Tobacco Mosaic Virus are?

    • DNA and RNA

    • RNA and Protein

    • DNA and Protein

    • Lipids and nucleic acids

    • Carbohydrates and nucleic acids

    Correct Answer
    A. RNA and Protein
    Explanation
    Tobacco Mosaic Virus is composed of RNA and protein. RNA is the genetic material of the virus, which carries the instructions for viral replication and protein synthesis. The protein component of the virus plays a crucial role in protecting the RNA and facilitating its entry into host cells. Together, RNA and protein make up the major components of Tobacco Mosaic Virus.

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  • 12. 

    Which of the following enzymes initiate chain synthesis on a template during replication? 

    • DNA pol I

    • DNA pol II

    • DNA pol III

    • RNA pol II

    • Primase

    Correct Answer
    A. Primase
    Explanation
    Primase is the correct answer because it is an enzyme that synthesizes RNA primers during DNA replication. These RNA primers serve as starting points for DNA synthesis by DNA polymerases. DNA pol I, DNA pol II, and DNA pol III are DNA polymerases that are involved in DNA replication but do not initiate chain synthesis. RNA pol II is an enzyme involved in transcription, not replication.

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  • 13. 

    Assume that a cross is made between tall and dwarf tobacco plants.  The F1 generation showed intermediate height, while the F2 generation showed a distribution of height ranging from tall to dwarf, like the original parents, and many heights between the extremes.  These data are consistent with the following mode of inheritance:

    • Multiple-factor inheritance

    • Alternation of generations

    • Codominance

    • Incomplete dominance

    • Hemizygosity

    Correct Answer
    A. Multiple-factor inheritance
    Explanation
    The F1 generation showing intermediate height suggests that there are multiple factors or genes involved in determining the height of tobacco plants. The F2 generation showing a distribution of heights ranging from tall to dwarf, like the original parents, and many heights between the extremes further supports the idea of multiple-factor inheritance. This means that the height of the tobacco plants is influenced by the combination of different genes or factors, rather than being determined by a single gene.

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  • 14. 

    Given the sequence 5’-AGTTACCTGA-3’ what would be the sequence of the complementary DNA strand?

    • 5’-TCAATGGACT-3’

    • 3’-AGTTACCTGA-5’

    • 5’-AGTTACCTGA-3’

    • 3’-TCAGGTAACT-5’

    • 5’-TCAGGTAACT-3’

    Correct Answer
    A. 5’-TCAGGTAACT-3’
    Explanation
    The complementary DNA strand is formed by pairing each nucleotide with its complementary base. In DNA, adenine (A) pairs with thymine (T), and cytosine (C) pairs with guanine (G). Therefore, in the given sequence 5’-AGTTACCTGA-3’, the complementary DNA strand would be 5’-TCAGGTAACT-3’.

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  • 15. 

    If the percentage of uracil in a double-stranded RNA molecule is 30%, the percentage of cytosine is?

    • 20%

    • 10%

    • 30%

    • 60%

    • 40%

    Correct Answer
    A. 20%
    Explanation
    In a double-stranded RNA molecule, the percentage of uracil and cytosine should be equal since they pair together. If the percentage of uracil is 30%, then the percentage of cytosine should also be 30%. Therefore, the answer of 20% is incorrect.

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  • 16. 

    In a double-stranded RNA molecule of 10,000 base pairs, the approximate number of complete turns is?

    • 3000

    • 15,000

    • 1000

    • 300

    • 30

    Correct Answer
    A. 1000
    Explanation
    In a double-stranded RNA molecule, the two strands are twisted around each other in a helical structure. The number of complete turns refers to the number of times the two strands wrap around each other in a full circle. In this case, the approximate number of complete turns in a 10,000 base pair RNA molecule is 1000. This means that the two strands wrap around each other 1000 times to form the helical structure.

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  • 17. 

    Hershey and Chase differentiated between DNA and protein by:

    • Labeling the DNA with 14Carbon, proteins with 3Hydrogen

    • Labeling the DNA with 35Sulfur, proteins with 32Phosphorous

    • Labeling the DNA with cesium, proteins with chloride

    • Labeling the DNA with 32Phosphorous, proteins with 35Sulfur

    • All of the above

    Correct Answer
    A. Labeling the DNA with 32Phosphorous, proteins with 35Sulfur
    Explanation
    Hershey and Chase differentiated between DNA and protein by labeling the DNA with 32Phosphorous and the proteins with 35Sulfur. This method allowed them to track the movement of these labeled molecules during the infection process. Since DNA contains phosphorous but not sulfur, and proteins contain sulfur but not phosphorous, they were able to determine that the genetic material of bacteriophages (viruses that infect bacteria) is DNA, not protein.

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  • 18. 

    What is the name for the statistical measure used to describe sample variability

    • Mean

    • Standard Error

    • Variance

    • Median

    • Covariance

    Correct Answer
    A. Variance
    Explanation
    Variance is the statistical measure used to describe sample variability. It quantifies how spread out the data points in a sample are from the mean. A higher variance indicates that the data points are more scattered, while a lower variance suggests that the data points are closer to the mean. Variance is calculated by taking the average of the squared differences between each data point and the mean.

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  • 19. 

    For a species with a diploid number of 18 chromosomes, how many chromosomes will be present in the somatic nuclei of individuals that are tetraploid?

    • 9

    • 27

    • 36

    • 18

    • 72

    Correct Answer
    A. 36
    Explanation
    When a species has a diploid number of 18 chromosomes, it means that each cell normally contains two sets of 18 chromosomes, totaling 36 chromosomes. In a tetraploid individual, there are four sets of chromosomes instead of the usual two. Therefore, the somatic nuclei of tetraploid individuals would contain four sets of 18 chromosomes, resulting in a total of 72 chromosomes.

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  • 20. 

    ________ results when chromatids fail to separate and move to opposite poles during anaphase. 

    • Nondisjunction

    • Transposition

    • Translocation

    • Duplication

    • Segregation

    Correct Answer
    A. Nondisjunction
    Explanation
    Nondisjunction is the correct answer because it refers to the failure of chromatids to separate and move to opposite poles during anaphase. This can result in an abnormal distribution of chromosomes in the daughter cells, leading to genetic disorders or abnormalities. Transposition, translocation, duplication, and segregation do not specifically refer to the failure of chromatids to separate during anaphase.

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  • 21. 

    The recombination frequency between three genes can be used to calculate

    • Actual distance between genes

    • Gene order

    • Number of tetrads

    • All of the above

    • None of the above

    Correct Answer
    A. Gene order
    Explanation
    The recombination frequency between three genes can be used to calculate the gene order. Recombination frequency is a measure of how often recombination occurs between two genes during meiosis. By determining the frequency of recombination between three genes, their relative positions on a chromosome can be inferred, allowing the determination of the gene order.

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  • 22. 

    The phenomenon of crossovers in nearby regions is called

    • Chiasma

    • Negative Interference

    • Reciprocal genetic exchange

    • Positive Interference

    • Mitotic recombination

    Correct Answer
    A. Chiasma
    Explanation
    Chiasma is the correct answer because it refers to the phenomenon of crossovers in nearby regions during the process of genetic recombination. During meiosis, homologous chromosomes exchange genetic material at specific points called chiasmata, resulting in the formation of recombinant chromosomes. This process leads to genetic diversity and plays a crucial role in evolution.

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  • 23. 

    The mating system in E.coli is known as?

    • Lysogeny

    • Lysis

    • Conjugation

    • Integration

    • Transduction

    Correct Answer
    A. Conjugation
    Explanation
    Conjugation is the correct answer because it refers to the process by which E. coli transfers genetic material to another bacterium through direct cell-to-cell contact. This process allows for the exchange of plasmids, which are small, circular pieces of DNA that can carry beneficial genes such as antibiotic resistance. Conjugation plays a crucial role in bacterial evolution and the spread of antibiotic resistance genes among bacterial populations.

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  • 24. 

    The transfer of bacterial genes from one cell to another requires?

    • F factor

    • F plasmids

    • F + cell

    • All of the above

    • None of the above

    Correct Answer
    A. All of the above
    Explanation
    The transfer of bacterial genes from one cell to another requires multiple factors, including the F factor, F plasmids, and F+ cells. The F factor is a piece of DNA that allows for the transfer of genetic material between bacterial cells. F plasmids are self-replicating circular pieces of DNA that contain the F factor and are responsible for transferring the F factor to other cells. F+ cells are bacterial cells that contain the F factor and are capable of transferring genetic material to F- cells. Therefore, all of these factors are necessary for the transfer of bacterial genes.

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  • 25. 

    Which of the following is a nonhistone protein found in chromatin? 

    • H1

    • HMG

    • H2A

    • H3

    Correct Answer
    A. HMG
    Explanation
    HMG (High Mobility Group) is a nonhistone protein found in chromatin. It functions as a structural protein that helps in the organization and stabilization of chromatin structure. HMG proteins are involved in various cellular processes such as transcription, DNA repair, and recombination. They bind to DNA and help in regulating gene expression by facilitating the binding of transcription factors and other proteins to DNA. Unlike histone proteins, which are also found in chromatin, HMG proteins do not participate in nucleosome formation but play a crucial role in maintaining chromatin architecture.

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  • 26. 

    Chromatin contains

    • A) DNA.

    • B) RNA.

    • C) protein.

    • D) both A and C

    • E) all of the above

    Correct Answer
    A. D) both A and C
    Explanation
    Chromatin is a complex structure found in the nucleus of cells that contains DNA and proteins. DNA is the genetic material that carries the instructions for the development and functioning of all living organisms. Proteins, on the other hand, play various roles in the organization and regulation of DNA. Therefore, the correct answer is D) both A and C, as chromatin contains both DNA and protein.

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  • 27. 

    Plasmids that can confer the ability to conjugate are known as?

    • F plasmids

    • Colicin plasmids

    • Bacteriophage

    • E. coli

    • R plasmids

    Correct Answer
    A. F plasmids
    Explanation
    F plasmids are known as plasmids that can confer the ability to conjugate. Conjugation is a mechanism of horizontal gene transfer in bacteria where genetic material is transferred between two bacterial cells through direct cell-to-cell contact. F plasmids contain the Fertility (F) factor which allows them to transfer their genetic material to recipient cells during conjugation. This transfer can lead to the acquisition of new traits or resistance to antibiotics. Therefore, F plasmids are specifically associated with the ability to conjugate.

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  • 28. 

    Which of the following codon pairs specify amino acids

    • AUG and GUG

    • UGA and AUG

    • AUG and UAA

    • UAA and UGA

    Correct Answer
    A. AUG and GUG
    Explanation
    The codon pairs AUG and GUG both specify the amino acid methionine. AUG is the start codon and is responsible for initiating protein synthesis, while GUG is an alternative codon that also codes for methionine. Therefore, both codon pairs can be used to specify the same amino acid.

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  • 29. 

    DNA synthesis in eukaryotes is?

    • Semi-conservative

    • Proceeds in a 3’ to 5’ direction

    • Involves a single protein

    • All of the above

    • None of the above

    Correct Answer
    A. Semi-conservative
    Explanation
    DNA synthesis in eukaryotes is semi-conservative. This means that during replication, each strand of the original DNA molecule serves as a template for the synthesis of a new complementary strand. As a result, each new DNA molecule formed contains one original strand and one newly synthesized strand. The other options, proceeding in a 3' to 5' direction and involving a single protein, are not accurate statements about DNA synthesis in eukaryotes.

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  • 30. 

                   actually synthesizes RNA, not DNA. 

    • DNA Pol I

    • Telomerase

    • Helicase

    • Primase

    • Gyrase

    Correct Answer
    A. Primase
    Explanation
    Primase is the correct answer because it is an enzyme that is responsible for synthesizing RNA primers during DNA replication. These RNA primers are necessary for DNA polymerase to initiate the synthesis of new DNA strands. Primase synthesizes short RNA sequences that serve as a starting point for DNA replication. It is important to note that primase synthesizes RNA, not DNA, as mentioned in the question.

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  • 31. 

                     contains an RNA molecule which is required for function.

    • DNA Pol I

    • Telomerase

    • Helicase

    • Primase

    • Gyrase

    Correct Answer
    A. Telomerase
    Explanation
    Telomerase is the correct answer because it contains an RNA molecule that is required for its function. Telomerase is an enzyme that adds repetitive nucleotide sequences to the ends of chromosomes, called telomeres. These telomeres protect the chromosomes from degradation and prevent the loss of important genetic information during DNA replication. Telomerase contains an RNA component that serves as a template for the addition of these telomere sequences, allowing the enzyme to extend the telomeres and maintain the integrity of the chromosomes.

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  • 32. 

                    relieves torsional stress.

    • DNA Pol I

    • Telomerase

    • Helicase

    • Primase

    • Gyrase

    Correct Answer
    A. Gyrase
    Explanation
    Gyrase is the correct answer because it is an enzyme that relieves torsional stress during DNA replication. Torsional stress occurs when the DNA double helix becomes overwound or underwound during the unwinding process. Gyrase helps to alleviate this stress by introducing negative supercoils into the DNA, allowing for smooth unwinding and replication to occur. This ensures that the DNA strands do not become tangled or damaged during the replication process.

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  • 33. 

                    opens the double helix for replication machinery.

    • DNA Pol I

    • Telomerase

    • Helicase

    • Primase

    • Gyrase

    Correct Answer
    A. Helicase
    Explanation
    Helicase is responsible for opening the double helix structure of DNA during replication. It does this by breaking the hydrogen bonds between the DNA strands, allowing the replication machinery to access the DNA template strands and synthesize new complementary strands. Helicase acts as a molecular motor, using energy from ATP hydrolysis to unwind the DNA helix and separate the strands. This process is crucial for DNA replication, as it provides the single-stranded DNA templates needed for DNA polymerases to synthesize new DNA strands.

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  • 34. 

                    removes RNA primer.

    • DNA Pol I

    • Telomerase

    • Helicase

    • Primase

    • Gyrase

    Correct Answer
    A. DNA Pol I
    Explanation
    DNA Pol I is responsible for removing RNA primers during DNA replication. RNA primers are short sequences of RNA that are synthesized by the enzyme Primase and serve as a starting point for DNA synthesis. Once DNA synthesis is complete, DNA Pol I replaces the RNA primers with DNA nucleotides, effectively removing them from the newly synthesized DNA strand. This process is essential for the completion of DNA replication and the production of a continuous DNA strand. Telomerase is involved in the replication of telomeres, Helicase unwinds the DNA double helix, Primase synthesizes RNA primers, and Gyrase relieves the tension caused by the unwinding of the DNA.

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  • 35. 

    Which of the following is not different between eukaryotic and prokaryotic transcription?

    • Splicing

    • 5' capping

    • Poly-adenylation

    • Termination

    • All are different

    Correct Answer
    A. All are different
    Explanation
    The correct answer is "all are different". This means that all of the options listed (splicing, 5' capping, poly-adenylation, and termination) are different between eukaryotic and prokaryotic transcription. In eukaryotes, splicing is required to remove introns from the pre-mRNA, 5' capping is the addition of a modified guanine nucleotide to the 5' end of the mRNA, poly-adenylation is the addition of a poly-A tail to the 3' end of the mRNA, and termination involves the recognition of specific termination sequences. In contrast, prokaryotic transcription does not involve splicing, 5' capping, or poly-adenylation, and termination occurs differently.

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  • 36. 

    The correct order of activity in replication is

    • Helicase, SSB, Pol III, primase, Pol I, ligase

    • Primase, helicase, SSB, Pol III, Pol I, ligase

    • SSB, helicase, primase, Pol III, Pol I, ligase

    • Helicase, SSB, primase, Pol I, Pol III, ligase

    • Helicase, SSB, primase, Pol III, Pol I, ligase

    Correct Answer
    A. Helicase, SSB, primase, Pol III, Pol I, ligase
    Explanation
    The correct order of activity in replication is helicase, SSB, primase, Pol III, Pol I, and ligase. Helicase is responsible for unwinding the DNA double helix, allowing the replication process to begin. Single-strand binding proteins (SSB) then bind to the single-stranded DNA to stabilize it and prevent re-annealing. Primase synthesizes RNA primers that are necessary for DNA polymerase III (Pol III) to initiate replication. Pol III is the main DNA polymerase responsible for synthesizing new DNA strands. Pol I then removes the RNA primers and replaces them with DNA. Finally, ligase joins the Okazaki fragments on the lagging strand and seals any remaining nicks in the DNA backbone.

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  • 37. 

    Before a prokaryotic mRNA can be translated, it must be modified by the 

    • Addition of a 5' cap.

    • Addition of a polyA tail

    • Removal of introns

    • All of the above

    • None of the above

    Correct Answer
    A. None of the above
    Explanation
    The correct answer is "none of the above". This is because prokaryotic mRNA does not undergo the same modifications as eukaryotic mRNA. Prokaryotic mRNA does not have a 5' cap or a polyA tail, and it also does not contain introns. Therefore, none of the modifications mentioned in the options are required for prokaryotic mRNA translation.

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  • 38. 

    Enzyme involved in transcription of tRNA

    • DNA Polymerase I

    • RNA Polymerase II

    • Aminoacyl transferase

    • RNA Polymerase III

    Correct Answer
    A. RNA Polymerase III
    Explanation
    RNA Polymerase III is the correct answer because it is the enzyme involved in the transcription of tRNA. RNA polymerases are responsible for synthesizing RNA molecules from a DNA template during transcription. RNA Polymerase III specifically transcribes small non-coding RNAs, including tRNA, which are essential for protein synthesis. DNA Polymerase I is involved in DNA replication and repair, while RNA Polymerase II is responsible for transcribing protein-coding genes. Aminoacyl transferase is not directly involved in transcription but plays a role in attaching amino acids to tRNA molecules during protein synthesis.

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  • 39. 

    Enzyme involved in transcription of mRNA

    • DNA Polymerase I

    • RNA Polymerase II

    • Aminoacyl transferase

    • RNA Polymerase III

    Correct Answer
    A. RNA Polymerase II
    Explanation
    RNA Polymerase II is the correct answer because it is the enzyme responsible for transcribing DNA into mRNA during the process of transcription. RNA Polymerase II recognizes specific DNA sequences called promoters and binds to them, initiating the synthesis of mRNA from the DNA template. This enzyme plays a crucial role in gene expression by transcribing the genetic information stored in DNA into a functional mRNA molecule, which can then be translated into proteins. DNA Polymerase I, Aminoacyl transferase, and RNA Polymerase III are not directly involved in the transcription of mRNA.

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  • 40. 

    DNA-dependent DNA polymerase

    • DNA Polymerase I

    • RNA Polymerase II

    • Aminoacyl transferase

    • RNA Polymerase III

    Correct Answer
    A. DNA Polymerase I
    Explanation
    DNA Polymerase I is the correct answer because it is an enzyme that is involved in DNA replication and repair. It is responsible for removing the RNA primers during DNA replication and replacing them with DNA nucleotides. DNA Polymerase I also has a 5' to 3' exonuclease activity, which allows it to proofread and correct errors in the newly synthesized DNA strand. Additionally, DNA Polymerase I has a role in DNA recombination and the repair of damaged DNA.

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  • 41. 

    Enzyme that charges tRNAs

    • DNA Polymerase I

    • RNA Polymerase II

    • Aminoacyl transferase

    • RNA Polymerase III

    Correct Answer
    A. Aminoacyl transferase
    Explanation
    Aminoacyl transferase is the enzyme responsible for charging tRNAs. Charging refers to the process of attaching the correct amino acid to the corresponding tRNA molecule, which is necessary for protein synthesis. DNA Polymerase I is involved in DNA replication, not tRNA charging. RNA Polymerase II and RNA Polymerase III are responsible for transcribing DNA into RNA, but they are not involved in charging tRNAs. Therefore, the correct answer is Aminoacyl transferase.

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  • 42. 

    Given the following sequence, how large (# amino acids) is the most likely eukaryotic polypeptide produced?                                       ACGAUGAGGAGGAUGGUCAAUGAUGCUGGCUGUUGAUAUUAACAU

    • 10

    • 7

    • 3

    • 4

    • 2

    Correct Answer
    A. 7
    Explanation
    The most likely eukaryotic polypeptide produced can be determined by identifying the start codon (AUG) and counting the number of codons (groups of three nucleotides) until a stop codon is encountered. In this sequence, the start codon is ACGAUGAGGAGGAUGGUC, and the stop codon is UAA. Counting the codons between the start and stop codons gives a total of 7 codons. Since each codon represents an amino acid, the most likely eukaryotic polypeptide produced would have 7 amino acids.

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  • 43. 

    If the sequence of a nontemplate strand of DNA is 5’-ACCGCATCCGAGTCAC-3’, what is the sequence of the primary product of transcription?

    • 3’-UGGCGUAGGCUCACUG-5’

    • 3’-TGGCGTAGGCTCACTG-5’

    • 5’-ACCGCAUCCGAGUCAC-3’

    • 5’-ACCGCATCCGAGTCAC-3’

    • None of the above

    Correct Answer
    A. 5’-ACCGCAUCCGAGUCAC-3’
    Explanation
    The correct answer is 5’-ACCGCAUCCGAGUCAC-3’. In transcription, the DNA sequence is used as a template to produce a complementary RNA molecule. The RNA molecule is synthesized in the 5’ to 3’ direction, matching the DNA sequence with the exception that thymine (T) is replaced with uracil (U). Therefore, the sequence of the primary product of transcription would be the same as the nontemplate strand of DNA, but with T replaced by U. In this case, the nontemplate strand is 5’-ACCGCATCCGAGTCAC-3’, and the correct answer, 5’-ACCGCAUCCGAGUCAC-3’, matches this pattern.

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  • 44. 

    The genetic code is said to be “degenerate” because 

    • There are more codons than amino acids.

    • There are more amino acids than codons.

    • Different organisms use different codons to encode the same amino acid.

    • Some codons specify more than one amino acid.

    Correct Answer
    A. There are more codons than amino acids.
    Explanation
    The genetic code is said to be "degenerate" because there are more codons than amino acids. This means that multiple codons can code for the same amino acid. For example, the amino acid leucine can be encoded by six different codons. This redundancy in the genetic code provides flexibility and robustness, as it allows for some variations or mutations in the DNA sequence without necessarily changing the resulting protein sequence.

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  • 45. 

    When one speaks of a 5' cap, one is usually describing?

    • Addition of a thymine to the 5’ end of an mRNA

    • Addition of a guanine to the 3’ end of an mRNA

    • Addition of a guanine to the 3’ end of an mRNA

    • Addition of a guanine to the 5’ end of an mRNA

    • None of the above

    Correct Answer
    A. Addition of a guanine to the 5’ end of an mRNA
    Explanation
    The correct answer is addition of a guanine to the 5’ end of an mRNA. The 5' cap is a modified guanine nucleotide that is added to the 5' end of the mRNA molecule during RNA processing. This modification helps protect the mRNA from degradation and is involved in various cellular processes such as mRNA export, translation initiation, and mRNA stability.

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  • 46. 

    If the sequence of an RNA molecule is 5’-GGCAUCGACG-3’, what is the sequence of the template strand of DNA?

    • 5’-GGCATCGACG-3’

    • 3’-GGCATCGACG-5’

    • 5’-CCGTAGCTGC-3’

    • 3’-CCGTAGCTGC-5’

    • None of the above

    Correct Answer
    A. 3’-CCGTAGCTGC-5’
    Explanation
    The sequence of the template strand of DNA is the complementary sequence to the RNA molecule. In RNA, adenine (A) pairs with uracil (U), cytosine (C) pairs with guanine (G), and vice versa in DNA. Therefore, the template strand of DNA would have the sequence 3’-CCGTAGCTGC-5’, which is the complementary sequence to the given RNA sequence.

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  • 47. 

    The primary structure of a protein is determined by

    • The sequence of amino acids

    • Hydrogen bonds formed between the components of the peptide linkage

    • A series of helical domains

    • Pleated sheets

    • Covalent bonds formed between fibroin residues

    Correct Answer
    A. The sequence of amino acids
    Explanation
    The primary structure of a protein refers to the specific sequence of amino acids that make up the protein chain. This sequence is determined by the genetic code and is crucial for the overall structure and function of the protein. The sequence of amino acids determines how the protein folds into its three-dimensional shape and how it interacts with other molecules in the body. Therefore, the correct answer is that the primary structure of a protein is determined by the sequence of amino acids.

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  • 48. 

    In the absence of glucose, the CAP protein binds to a DNA sequence adjacent to the promoter of the lac operon.  Binding of CAP helps RNA polymerase to bind to the promoter and allows for a high level of transcription of the lac operon.  Regulation of the lac operon by the CAP protein is an example of

    • Negative regulation

    • Positive regulation

    • Induction

    • All allosteric effect

    • Constitutive expression

    Correct Answer
    A. Positive regulation
    Explanation
    The binding of the CAP protein to the DNA sequence adjacent to the promoter of the lac operon in the absence of glucose helps RNA polymerase to bind to the promoter, resulting in a high level of transcription of the lac operon. This indicates that the CAP protein positively regulates the lac operon by enhancing the transcription process.

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  • 49. 

    Most genes in prokaryotes and eukaryotes are regulated primarily at which level of expression?

    • Transcription

    • MRNA stability

    • RNA processing

    • Protein activation

    • Translation

    Correct Answer
    A. Transcription
    Explanation
    Genes in both prokaryotes and eukaryotes are primarily regulated at the level of transcription. This is because transcription is the process by which the genetic information in DNA is used to synthesize RNA molecules, which then serve as templates for protein synthesis. By controlling the rate of transcription, cells can regulate the amount of RNA produced and therefore the amount of protein that is ultimately synthesized. This allows cells to respond to changing environmental conditions and to regulate gene expression in a precise and coordinated manner.

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Quiz Review Timeline (Updated): Apr 29, 2024 +

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  • Current Version
  • Apr 29, 2024
    Quiz Edited by
    ProProfs Editorial Team
  • Dec 06, 2010
    Quiz Created by
    ProfGentle

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