Biology 101 Test #4

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Biology Quizzes & Trivia

Questions and Answers
  • 1. 

    Homologous chromosomes align on the metaphase plate during ______. Sister chromatids align along the metaphse plate during ________.

    • A.

      Metaphase II, Prophase II

    • B.

      Metaphase I, Metaphase II

    • C.

      Anaphase I, Anaphase II

    • D.

      Prophase II, Metaphase II

    Correct Answer
    B. Metaphase I, Metaphase II
    Explanation
    During metaphase I of meiosis, homologous chromosomes align on the metaphase plate. This is the stage where these pairs of chromosomes, one from each parent, come together and line up side by side. On the other hand, during metaphase II, sister chromatids align along the metaphase plate. This occurs after the homologous chromosomes have separated in anaphase I and each chromosome has replicated into two sister chromatids. These sister chromatids align individually on the metaphase plate in preparation for their separation in anaphase II.

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  • 2. 

    Which event is unique to mitosis, but not meiosis?

    • A.

      Formation of tetrads

    • B.

      Crossing over of non-sister chromatids

    • C.

      Chromosomes line up on the metaphase plate independently, not in pairs

    • D.

      Sister chromatids separating during anaphase

    Correct Answer
    C. Chromosomes line up on the metaphase plate independently, not in pairs
    Explanation
    In mitosis, chromosomes line up on the metaphase plate independently, not in pairs. This is a unique event to mitosis because in meiosis, chromosomes line up on the metaphase plate in pairs. In meiosis, homologous chromosomes pair up during prophase I to form tetrads, and crossing over of non-sister chromatids occurs during this stage as well. Sister chromatids separating during anaphase is a common event in both mitosis and meiosis.

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  • 3. 

    Meiosis and cytokinesis result in the production of...?

    • A.

      2 diploid daughter cells

    • B.

      4 diploid daughter cells

    • C.

      2 haploid daughter cells

    • D.

      4 haploid daughter cells

    Correct Answer
    D. 4 haploid daughter cells
    Explanation
    Meiosis is a type of cell division that occurs in reproductive cells and results in the formation of haploid daughter cells. During meiosis, the parent cell undergoes two rounds of division, resulting in four daughter cells. These daughter cells are haploid, meaning they contain half the number of chromosomes as the parent cell. Cytokinesis, on the other hand, is the process of dividing the cytoplasm of the cell. Therefore, the correct answer is 4 haploid daughter cells, as meiosis and cytokinesis together lead to the production of four haploid daughter cells.

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  • 4. 

    Crossing over events occur between...

    • A.

      Sister chromatids

    • B.

      Non-homologous chromosomes

    • C.

      Non-sister chromatids

    • D.

      Identical strands of DNA

    Correct Answer
    C. Non-sister chromatids
    Explanation
    Crossing over events occur between non-sister chromatids. During crossing over, genetic material is exchanged between non-sister chromatids of homologous chromosomes. This process results in genetic recombination and contributes to genetic diversity. Non-sister chromatids are different chromatids of the same homologous pair, while sister chromatids are identical copies of each other. Therefore, crossing over events do not occur between sister chromatids, but rather between non-sister chromatids.

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  • 5. 

    Why is meiosis a necessary component to eukaryote sex?

    • A.

      Since the maternal and paternal gametes fuse, they must be produced as diploid cells by meiosis, or the ploidy number will be cut in half each generation.

    • B.

      A large number of gametes are needed, so producing four daughter cells is more efficient.

    • C.

      Since the maternal and paternal gametes fuse, they must be produced as haploid cells by meiosis, or the ploidy number will double each generation.

    • D.

      Crossing over and independent assortment during meiosis II create a high degree of genetic variability, ensuring that at least some progeny will be successful.

    Correct Answer
    C. Since the maternal and paternal gametes fuse, they must be produced as haploid cells by meiosis, or the ploidy number will double each generation.
    Explanation
    Meiosis is necessary in eukaryote sex because it produces haploid cells, which have half the number of chromosomes as the parent cells. When the maternal and paternal gametes fuse during fertilization, the resulting zygote will have the correct number of chromosomes, maintaining the ploidy level. If meiosis did not occur and the gametes were not haploid, the ploidy number would double with each generation, resulting in an unbalanced number of chromosomes and genetic abnormalities. Therefore, meiosis ensures the correct ploidy level and genetic stability in eukaryote sex.

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  • 6. 

    Meiosis and sexual reproduction increase diversity because...

    • A.

      They are archaic processes.

    • B.

      It allows for populations to adapt to environmental changes.

    • C.

      They produce offspring extremely quickly.

    • D.

      They will almost always have different offspring arise.

    Correct Answer
    D. They will almost always have different offspring arise.
    Explanation
    Meiosis and sexual reproduction increase diversity because they involve the shuffling and recombination of genetic material. During meiosis, the chromosomes in a cell undergo genetic recombination, resulting in the formation of gametes with unique combinations of genetic information. When these gametes fuse during sexual reproduction, they create offspring that inherit a combination of traits from both parents, leading to increased genetic diversity within a population. This diversity is advantageous as it allows for populations to adapt to environmental changes, increasing their chances of survival and evolution.

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  • 7. 

    Which event distinguishes anaphase from anaphase II?

    • A.

      Anaphase produces haploid daughter cells and anaphase II results in diploid daughter cells.

    • B.

      Sister chromatids separated in anaphase II are different while they are identical in anaphase.

    • C.

      The ploidy level will be reduced in anaphase II.

    • D.

      Spindle fibers separate the kinetochores in chromosomes in anaphase.

    Correct Answer
    B. Sister chromatids separated in anaphase II are different while they are identical in anaphase.
    Explanation
    In anaphase, sister chromatids are separated and pulled to opposite ends of the cell, resulting in the formation of two identical daughter cells. However, in anaphase II, sister chromatids are separated again, but this time they are no longer identical. This is because during meiosis, genetic recombination occurs during prophase I, which shuffles and exchanges genetic material between homologous chromosomes. As a result, the sister chromatids in anaphase II have different combinations of genetic information compared to their counterparts in anaphase.

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  • 8. 

    Crossing over events occur during which phase of meiosis?

    • A.

      Prophase I

    • B.

      Metaphase I

    • C.

      Anaphase II

    • D.

      Prophase II

    Correct Answer
    A. Prophase I
    Explanation
    Crossing over events occur during Prophase I of meiosis. This phase is characterized by the pairing of homologous chromosomes, which then undergo a process called synapsis. During synapsis, crossing over occurs, where sections of genetic material are exchanged between the paired chromosomes. This genetic recombination increases genetic diversity and ensures the proper distribution of genetic material during the subsequent stages of meiosis. Therefore, Prophase I is the correct phase for crossing over events to occur.

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  • 9. 

    How does metaphase in meiosis I and meiosis II differ?

    • A.

      Sister chromatids are on the metaphase plate in meiosis I and tetrads are on the metaphase plate in meiosis II.

    • B.

      Homologous chromosomes line up in meiosis I and duplicated chromosomes line up in meiosis II.

    • C.

      All chromatids are the exact same in meiosis I and differ in meiosis II due to independent assortment.

    • D.

      The ploidy level remains the same in meiosis I but will be reduced in meiosis II.

    Correct Answer
    B. Homologous chromosomes line up in meiosis I and duplicated chromosomes line up in meiosis II.
    Explanation
    In metaphase of meiosis I, homologous chromosomes line up on the metaphase plate. This is because meiosis I involves the pairing of homologous chromosomes, which are similar in size and carry genes for the same traits. On the other hand, in metaphase of meiosis II, duplicated chromosomes line up on the metaphase plate. This is because meiosis II involves the separation of sister chromatids, which are identical copies of each other formed during DNA replication. Therefore, the correct answer is that homologous chromosomes line up in meiosis I and duplicated chromosomes line up in meiosis II.

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  • 10. 

    In the figure, in which area do microtubules attach?

    • A.

      A

    • B.

      B

    • C.

      C

    • D.

      D

    • E.

      E

    Correct Answer
    C. C
    Explanation
    Microtubules attach in area C.

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  • 11. 

    A reduction division occurs during...

    • A.

      Meiosis but not mitosis.

    • B.

      Mitosis but not meiosis.

    • C.

      Both mitosis and meiosis.

    • D.

      Neither mitosis or meiosis.

    Correct Answer
    A. Meiosis but not mitosis.
    Explanation
    During meiosis, a reduction division occurs where the number of chromosomes is halved. This is important for sexual reproduction as it produces gametes with half the number of chromosomes as the parent cell. On the other hand, mitosis is a process of cell division that results in two identical daughter cells with the same number of chromosomes as the parent cell. Therefore, a reduction division only occurs during meiosis, not mitosis.

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  • 12. 

    Even though the human ovary at birth contains several hundred thousand oocyte-containing follicles, we can say that each is likely to be genetically unique. Why?

    • A.

      Independent assortment of chromosomes during anaphase I creates 2^23 (greater than 8 million) possible combinations of chromosomes, plus there is added variety from crossover events.

    • B.

      With high DNA mutation rates, each oocyte is likely to carry at least one unique base-pair mutation.

    • C.

      Random deletions of chromosome arms during chiasmata formation result in 2^46 different possible combinations of deletions.

    • D.

      The ovary lacks DNA damage checkpoints for gametes produced during mitosis.

    Correct Answer
    A. Independent assortment of chromosomes during anaphase I creates 2^23 (greater than 8 million) possible combinations of chromosomes, plus there is added variety from crossover events.
    Explanation
    The correct answer is that the independent assortment of chromosomes during anaphase I creates a large number of possible combinations of chromosomes, which results in genetic uniqueness. Additionally, crossover events during meiosis further contribute to genetic variation. This means that each oocyte is likely to have a different combination of chromosomes, making them genetically unique.

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  • 13. 

    How many of your chromosomes are exactly the same as those of your maternal grandmother?

    • A.

      None. Any chromosome from my grandmother become recombined with a chromosome from my grandfather in my mother's gonad during meiosis II of gamete formation.

    • B.

      About 11 or 12 (1/4 of the total) since I have four grandparents.

    • C.

      None. Any chromosome from my grandmother become recombined with a chromosome from my grandfather in my mother's gonad during meiosis I of gamete formation.

    • D.

      Anywhere from 0 to 23 of my chromosomes may be those from my maternal grandmother - there is no way to predict.

    Correct Answer
    C. None. Any chromosome from my grandmother become recombined with a chromosome from my grandfather in my mother's gonad during meiosis I of gamete formation.
    Explanation
    During meiosis I of gamete formation, the chromosomes from the maternal grandmother and paternal grandfather undergo recombination, resulting in new combinations of genetic material. Therefore, none of the chromosomes in an individual will be exactly the same as those of their maternal grandmother.

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  • 14. 

    Which of the following events is not mediated by microtubules?

    • A.

      Separation of chromosomes

    • B.

      Lengthening of newly formed daughter cells

    • C.

      Formation of the cleavage furrow

    • D.

      Alignment of the chromosomes on the metaphase plate

    Correct Answer
    C. Formation of the cleavage furrow
    Explanation
    The formation of the cleavage furrow is not mediated by microtubules. Microtubules play a crucial role in several cellular processes, including the separation of chromosomes during cell division, the alignment of chromosomes on the metaphase plate, and the lengthening of newly formed daughter cells. However, the formation of the cleavage furrow is primarily mediated by actin filaments, which contract and pinch the cell membrane, leading to the division of the cell into two daughter cells.

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  • 15. 

    Almost all life cycles will have haploid and diploid phases. The transition from haploid to diploid occurs when?

    • A.

      After DNA replication during the S phase of the cell cycle.

    • B.

      Once the origin of replication is complete and the septum has formed.

    • C.

      During crossing over in prophase I of meiosis.

    • D.

      When gametes fuse during fertilization.

    Correct Answer
    D. When gametes fuse during fertilization.
    Explanation
    The correct answer is when gametes fuse during fertilization. This is because the fusion of gametes, which are haploid cells, results in the formation of a zygote, which is a diploid cell. In most life cycles, including sexual reproduction in animals and plants, the fusion of gametes is the key event that leads to the transition from the haploid phase to the diploid phase. This is because the zygote contains two sets of chromosomes, one from each parent, and will undergo cell division to develop into a multicellular organism.

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  • 16. 

    During evolution, what four new features had to evolve to change a primitive mitosis machinery into one that could support meiosis?

    • A.

      Pairing of homologous chromosomes, crossover, sister chromatids staying together in anaphase II, and supression of DNA replication before meiosis I.

    • B.

      Pairing of homologous chromosomes, crossover, sister chromatids separating during anaphase II, and kinetochore microtubules.

    • C.

      Pairing of sister chromatids, crossover between sister chromatids, centrosomes containing centrioles, and chiasmata connecting the kinetochores to the centrosome.

    • D.

      Pairing of homologous chromosomes, crossover, sister chromatids staying together in anaphase I, and suppression of DNA replication during interphase between meiosis I and II.

    Correct Answer
    D. Pairing of homologous chromosomes, crossover, sister chromatids staying together in anaphase I, and suppression of DNA replication during interphase between meiosis I and II.
    Explanation
    The correct answer is the fourth option. During evolution, four new features had to evolve to change a primitive mitosis machinery into one that could support meiosis. These features include pairing of homologous chromosomes, crossover, sister chromatids staying together in anaphase I, and suppression of DNA replication during interphase between meiosis I and II. These changes are necessary for the proper segregation of genetic material and the generation of genetic diversity during meiosis.

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  • 17. 

    Humans, cats, elephants, and birds fall into what category of life cycles?

    • A.

      Haploid-dominant

    • B.

      Alternation of generations

    • C.

      Binary fission

    • D.

      Diploid-dominant

    Correct Answer
    D. Diploid-dominant
    Explanation
    The correct answer is diploid-dominant because humans, cats, elephants, and birds all have diploid cells, meaning they have two sets of chromosomes in their cells. In diploid-dominant life cycles, the majority of the organism's life is spent in the diploid phase, where the organism has two sets of chromosomes. This is in contrast to haploid-dominant life cycles, where the majority of the organism's life is spent in the haploid phase with one set of chromosomes.

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  • 18. 

    Which pair of chromosomes in the figure may not be homologous?

    • A.

      Pair 1

    • B.

      Pair 8

    • C.

      Pair 13

    • D.

      Pair 23

    Correct Answer
    D. Pair 23
  • 19. 

    In the figure, chiasmata are pulled apart in what phase?

    • A.

      Prophase I

    • B.

      Prophase II

    • C.

      Metaphase I

    • D.

      Metaphase II

    • E.

      Anaphase I

    • F.

      Anaphase II

    Correct Answer
    E. Anaphase I
    Explanation
    During Anaphase I of meiosis, the chiasmata, which are points of crossing over between homologous chromosomes, are pulled apart. This phase follows the alignment of homologous chromosomes in Metaphase I and precedes the separation of sister chromatids in Telophase I. Therefore, Anaphase I is the correct answer as it specifically describes the phase where chiasmata are separated.

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  • 20. 

    Which set of chromosomes is the most similar genetically?

    • A.

      Homologous chromosomes

    • B.

      The X and Y chromosome

    • C.

      Sister chromatids

    • D.

      A grandparent's chromosome 16 and the one which was passed down to his grandson

    Correct Answer
    A. Homologous chromosomes
    Explanation
    Homologous chromosomes are the most similar genetically because they contain the same genes in the same order, although they may have different alleles. They pair up during meiosis and undergo genetic recombination, which increases genetic diversity. The X and Y chromosomes determine the sex of an individual and have different genetic content. Sister chromatids are identical copies of a chromosome formed during DNA replication. A grandparent's chromosome 16 and the one passed down to his grandson may have similar genetic material, but they are not necessarily the most similar genetically as they can undergo genetic recombination during meiosis.

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  • 21. 

    If mitosis and meiosis are similar processes, why doesn't mitosis add genetic diversity?

    • A.

      Mitosis lacks crossing over between homologous chromosomes and since both chromosomes are partitioned to daughters, also lacks independent assortment.

    • B.

      DNA damage checkpoints are more vigilant during mitosis than during meiosis.

    • C.

      The chromosomes are less exposed to damaging UV rays during mitosis than during meiosis.

    • D.

      Mitosis does add genetic diversity by similar mechanisms, but the results are less apparent since an individual organism doesn't arise from the mitotic daughter cells.

    Correct Answer
    A. Mitosis lacks crossing over between homologous chromosomes and since both chromosomes are partitioned to daughters, also lacks independent assortment.
    Explanation
    Mitosis does not add genetic diversity because it lacks crossing over between homologous chromosomes and independent assortment. During mitosis, both chromosomes are partitioned to the daughter cells without any exchange of genetic material, resulting in identical copies of the original cell. This means that the genetic information remains the same and no new combinations of alleles are formed. In contrast, meiosis involves crossing over and independent assortment, leading to the creation of genetically diverse gametes.

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  • 22. 

    Which of these is not a main category of life cycles in multicellular organisms?

    • A.

      Haploid-dominant

    • B.

      Alternation of generations

    • C.

      Binary fission

    • D.

      Diploid-dominant

    Correct Answer
    C. Binary fission
    Explanation
    Binary fission is not a main category of life cycles in multicellular organisms. Binary fission is a form of asexual reproduction found in single-celled organisms, such as bacteria and protozoa. In binary fission, the organism simply divides into two identical daughter cells. On the other hand, multicellular organisms have more complex life cycles that involve processes like haploid-dominant (where the haploid stage is dominant), alternation of generations (where there are alternating haploid and diploid stages), and diploid-dominant (where the diploid stage is dominant).

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  • 23. 

    For a diploid cell (2n = 4), the number of chromosomes in each daughter cell at the end of meiosis II is ____; the number of chromosomes in each daughter cell at the completion of mitosis is ____.

    • A.

      2;2

    • B.

      2;4

    • C.

      4;4

    • D.

      4;8

    Correct Answer
    B. 2;4
    Explanation
    During meiosis II, the diploid cell undergoes another round of division resulting in the formation of four haploid daughter cells. Since the original cell had 4 chromosomes, each daughter cell will have 2 chromosomes.

    On the other hand, during mitosis, the diploid cell divides into two identical diploid daughter cells. Therefore, each daughter cell will have the same number of chromosomes as the original cell, which is 4 in this case.

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  • 24. 

    You are engaged in a project to breed new color variants in snakes. A certain chromosome carries three genes that all affect pigmentation. Two of the genes,  A and B are near each other on the chromosome, but gene C is further away. You are hoping to maintain the versions of the three genes from the maternal chromosome in some of the grandchildren. What would you predict for the behavior of these genes during crossover in meiosis?

    • A.

      All three genes are equally likely to be separated by a crossover event, since these are randomly distributed.

    • B.

      The maternal chromosome will be transmitted as it is to half of the progeny.

    • C.

      The maternal chromosome will be transmitted as it is to all of the progeny.

    • D.

      Genes A and B are likely to stay together, but they are more likely to become separated from gene C since a crossover is more likely to occur in the longer space between them.

    Correct Answer
    D. Genes A and B are likely to stay together, but they are more likely to become separated from gene C since a crossover is more likely to occur in the longer space between them.
    Explanation
    During crossover in meiosis, genes A and B are likely to stay together as they are near each other on the chromosome. However, they are more likely to become separated from gene C, which is further away from them. This is because crossovers are more likely to occur in the longer space between genes A and B and gene C. Therefore, the correct answer is that genes A and B are likely to stay together, but they are more likely to become separated from gene C.

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  • 25. 

    Sexual reproduction is advantageous over asexual reproduction when...

    • A.

      Organisms live a solitary lifestyle

    • B.

      When the parent organism have been successful in their habitat

    • C.

      Mutation rates are high

    • D.

      The environment is changing

    Correct Answer
    D. The environment is changing
    Explanation
    Sexual reproduction is advantageous over asexual reproduction when the environment is changing because it allows for genetic variation and diversity in offspring. This increased genetic diversity provides a greater chance for some individuals to possess traits that are better suited for the changing environment. In contrast, asexual reproduction produces offspring that are genetically identical to the parent organism, limiting their ability to adapt to new conditions. Therefore, sexual reproduction is more advantageous in a changing environment as it promotes adaptation and survival.

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  • 26. 

    What is the complement of 5׳ TCATTAACGG 3׳

    • A.

      5׳ TCATTAACGG 3

    • B.

      5׳ AGTAATTGCC 3

    • C.

      3׳ AGTAATTGCC 5׳

    • D.

      3׳ TCATTAACGG 5׳

    Correct Answer
    C. 3׳ AGTAATTGCC 5׳
    Explanation
    The complement of a DNA sequence is formed by replacing each nucleotide with its complementary base. In this case, the original sequence is 5׳ TCATTAACGG 3׳. The complementary base pairs are A-T and G-C. Therefore, the complement of the sequence would be 3׳ AGTAATTGCC 5׳, where each nucleotide has been replaced with its complementary base.

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  • 27. 

    Chargaff’s rule is supported by what statement?

    • A.

      A + T and C + G

    • B.

      A + C and T + G

    • C.

      A + G = T + C

    • D.

      A - T + G - C

    Correct Answer
    C. A + G = T + C
    Explanation
    Chargaff's rule states that in DNA, the amount of adenine (A) is equal to the amount of thymine (T), and the amount of cytosine (C) is equal to the amount of guanine (G). This is supported by the statement "A + G = T + C", which shows that the sum of the amounts of A and G is equal to the sum of the amounts of T and C. This statement aligns with the base pairing rules in DNA, where A pairs with T and C pairs with G.

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  • 28. 

    If a DNA sample from an unknown organism has telomerase, the organism

    • A.

      Is bacteria.

    • B.

      Is prokaryotic.

    • C.

      Is eukaryotic.

    • D.

      Has circular DNA.

    Correct Answer
    C. Is eukaryotic.
    Explanation
    The presence of telomerase in a DNA sample indicates that the organism is eukaryotic. Telomerase is an enzyme that helps in the replication and maintenance of telomeres, which are protective caps at the ends of chromosomes. Telomerase is mainly found in eukaryotic cells, where it helps to prevent the shortening of telomeres during DNA replication. Bacteria, on the other hand, do not possess telomerase and have different mechanisms for replicating their DNA. Therefore, the correct answer is that the organism is eukaryotic.

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  • 29. 

    A mutation causes a codon to change from UAC to UAU, both of which specify tyrosine. This is an example of

    • A.

      A frameshift mutation.

    • B.

      Missense mutation.

    • C.

      Silent mutation.

    • D.

      Nonsense mutation.

    Correct Answer
    C. Silent mutation.
    Explanation
    A silent mutation is a type of mutation that does not result in a change in the amino acid sequence of a protein. In this case, the mutation changes the codon from UAC to UAU, but both codons still specify the same amino acid, which is tyrosine. Therefore, this is an example of a silent mutation.

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  • 30. 

    In the disease sickle cell anemia, ACT (Thr)– CCT (Pro)– GAG (Glu) – GAG (Glu) is replaced by ACT (Thr) – CCT (Pro)– GTG (Val) – GAG (Glu). This represents a

    • A.

      Silent mutation.

    • B.

      Missense mutation.

    • C.

      Nonsense mutation.

    • D.

      Frameshift mutation.

    Correct Answer
    B. Missense mutation.
    Explanation
    The given sequence shows a single nucleotide substitution, where the codon GAG (Glu) is changed to GTG (Val). This results in the replacement of the amino acid glutamic acid (Glu) with valine (Val) in the protein. This type of mutation, where a single amino acid is changed, is known as a missense mutation.

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  • 31. 

    What is the correct order of DNA compaction in eukaryotes?

    • A.

      Helix, chromatin fiber, nucleosomes, duplicated chromosome, condensation of chromatin.

    • B.

      Condensation of chromatin, nucleosomes, chromatin fiber, helix, duplicated chromosome.

    • C.

      Chromatin fiber, condensation of chromatin, duplicated chromosome, nucleosome, helix.

    • D.

      Helix, nucleosome, chromatin fiber, condensation of chromatin, duplicated chromosome.

    Correct Answer
    D. Helix, nucleosome, chromatin fiber, condensation of chromatin, duplicated chromosome.
    Explanation
    In eukaryotes, DNA is first organized into a double helix structure. It then undergoes compaction by wrapping around histone proteins to form nucleosomes. These nucleosomes further condense to form a chromatin fiber. The condensation of chromatin leads to the formation of a duplicated chromosome, which consists of two identical copies of the DNA molecule. Therefore, the correct order of DNA compaction in eukaryotes is helix, nucleosome, chromatin fiber, condensation of chromatin, and duplicated chromosome.

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  • 32. 

    The DNA double helix model was proposed by

    • A.

      Avery.

    • B.

      Franklin.

    • C.

      Hershey and Chase.

    • D.

      Watson and Crick.

    Correct Answer
    D. Watson and Crick.
    Explanation
    Watson and Crick proposed the DNA double helix model. Their groundbreaking discovery in 1953 revolutionized our understanding of DNA structure and its role in heredity. They built upon the work of other scientists, such as Rosalind Franklin, who provided crucial X-ray crystallography data, to develop their model. Watson and Crick's model showed that DNA consists of two intertwined strands that form a helical structure, with specific base pairing rules. This model provided the foundation for further research in genetics and molecular biology.

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  • 33. 

    Frameshift mutations occur when

    • A.

      Bases are added or deleted.

    • B.

      There is an amino acid substitution.

    • C.

      There is a transition substitution.

    • D.

      Transversion substitution.

    Correct Answer
    A. Bases are added or deleted.
    Explanation
    Frameshift mutations occur when bases are added or deleted during DNA replication. This disrupts the reading frame of the genetic code, resulting in a shift in the grouping of codons. As a result, the entire sequence of amino acids encoded by the mutated gene is altered, leading to a non-functional or partially functional protein. A frameshift mutation can have significant consequences on the structure and function of the protein, often resulting in genetic disorders or diseases.

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  • 34. 

    What happens when a dideoxynucleotide is added a test tube?

    • A.

      Strand replication continues

    • B.

      Strand replication is unaffected

    • C.

      Strand replication is stopped

    • D.

      Strand replication slows down

    Correct Answer
    C. Strand replication is stopped
    Explanation
    When a dideoxynucleotide is added to a test tube, it acts as a chain terminator during DNA replication. Dideoxynucleotides lack the 3'-OH group needed for the formation of a phosphodiester bond, which is necessary for the addition of the next nucleotide in the growing DNA strand. Therefore, when a dideoxynucleotide is incorporated into the growing DNA strand, it prevents further elongation, leading to the termination of strand replication. This is why the correct answer is that strand replication is stopped.

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  • 35. 

    After centrifuging DNA that was developed in the Meselson-Stahl experiment, which type of DNA forms a band at the bottom of the tube?

    • A.

      DNA with 14N

    • B.

      DNA with 15N

    • C.

      DNA with equal amounts of 14N and 15N

    • D.

      Option 4

    Correct Answer
    B. DNA with 15N
    Explanation
    After centrifuging DNA that was developed in the Meselson-Stahl experiment, the DNA with 15N forms a band at the bottom of the tube. This is because the Meselson-Stahl experiment involved labeling DNA with heavy nitrogen isotopes, such as 15N. After multiple rounds of DNA replication, the DNA with 15N becomes denser and forms a band at the bottom of the tube during centrifugation. DNA with 14N and DNA with equal amounts of 14N and 15N would not be as dense and would form bands at higher positions in the tube.

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  • 36. 

    Amy and Bob work at the same biotech company. Amy is trying to develop telomerase activator drugs, and Bob is trying to develop telomerase inhibitor drugs. Why?

    • A.

      Amy wants a cure for warts, and Bob wants a therapy for heart disease.

    • B.

      Amy wants a cure for diabetes, and Bob wants a cure for MRSA (Methicillin-resistant Staphylococcus aureus) infections.

    • C.

      Amy wants a therapy for skin cancer, and Bob wants a new fertility treatment

    • D.

      Amy wants a therapy for the effects of aging, and Bob wants a cure for cancer.

    Correct Answer
    D. Amy wants a therapy for the effects of aging, and Bob wants a cure for cancer.
    Explanation
    Amy wants a therapy for the effects of aging because telomerase activator drugs can potentially slow down the aging process by lengthening telomeres, which are protective caps at the ends of chromosomes. On the other hand, Bob wants a cure for cancer because telomerase inhibitor drugs can inhibit the activity of telomerase, which is often overexpressed in cancer cells and allows them to divide indefinitely. By inhibiting telomerase, Bob hopes to stop the uncontrolled growth of cancer cells and ultimately find a cure for cancer.

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  • 37. 

    One of your young pediatric patients was discovered to have a mutation in a gene involved in nucleotide excision repair. What is the best recommendation you can make to the parents?

    • A.

      Discourage her from using tanning beds when she gets older, and put sunscreen on when she will be out in the summer sun for more than a few hours.

    • B.

      She should eat lots of fresh fruits and vegetables so her body can produce extra DNA on a regular basis.

    • C.

      Be sure she wears sun protective gear and sunscreen every day, avoids sunlight, and takes a Vitamin D supplement.

    • D.

      Discourage her from smoking cigarettes when she is older.

    Correct Answer
    C. Be sure she wears sun protective gear and sunscreen every day, avoids sunlight, and takes a Vitamin D supplement.
    Explanation
    The best recommendation for the parents is to ensure that the child wears sun protective gear and sunscreen every day, avoids sunlight, and takes a Vitamin D supplement. This is because the child has a mutation in a gene involved in nucleotide excision repair, which is responsible for repairing DNA damage caused by UV radiation. By taking these precautions, the child can reduce their risk of developing skin cancer and other UV-related health issues.

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  • 38. 

    Why are hydrogen bonds holding DNA bases together instead of covalent bonds?

    • A.

      Hydrogen bonds are stronger than covalent bonds

    • B.

      Hydrogen bonds are easier to break allowing for DNA copying of a template

    • C.

      Enzymes cannot break covalent bonds

    • D.

      Covalent bonds cannot attach purines and pyrimidines

    Correct Answer
    B. Hydrogen bonds are easier to break allowing for DNA copying of a template
    Explanation
    Hydrogen bonds are easier to break allowing for DNA copying of a template. This is because hydrogen bonds are weaker compared to covalent bonds. During DNA replication, the two strands of the DNA double helix need to separate so that each strand can serve as a template for the synthesis of a new complementary strand. If the bonds holding the bases together were strong covalent bonds, it would be difficult to separate the strands. However, hydrogen bonds can be easily broken, allowing for the separation of the DNA strands and the copying of the template strand.

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  • 39. 

    An experiment started with 15N15N DNA. After two generations in 14N medium, E. coli will have

    • A.

      25% 14N14N and 75% 15N15N

    • B.

      50% 14N14N and 50% 15N15N

    • C.

      75% 15N14N and 25% 14N14N

    • D.

      50% 15N14N and 50% 14N14N

    Correct Answer
    D. 50% 15N14N and 50% 14N14N
    Explanation
    After two generations in 14N medium, the E. coli will have a mixture of 15N14N and 14N14N DNA. This is because during DNA replication, the original 15N15N DNA will be separated into two strands, with one strand containing 15N and the other containing 14N. As the bacteria continue to divide and replicate their DNA, the 15N14N and 14N14N DNA will be passed on to subsequent generations, resulting in a 50% distribution of each type of DNA.

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  • 40. 

    You cannot see DNA with a standard microscope. So how did scientists like Hershey, Chase, Meselson and Stahl follow the DNA in their classic experiments?

    • A.

      Electron microscopy

    • B.

      Fluorescent labeling with transgenic GFP

    • C.

      Radioactive labeling

    • D.

      Covalent linkage to beads in a column

    Correct Answer
    C. Radioactive labeling
    Explanation
    Scientists like Hershey, Chase, Meselson, and Stahl were able to follow the DNA in their experiments by using radioactive labeling. This involves incorporating radioactive isotopes into the DNA molecules, which can then be detected using specialized equipment. By tracking the radioactive markers, scientists were able to observe the movement and behavior of DNA in various experiments, even though it cannot be seen directly with a standard microscope.

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  • 41. 

    Bacteria’s genetic material is

    • A.

      Single stranded and linear.

    • B.

      Double stranded and linear.

    • C.

      Single stranded and circular.

    • D.

      Doubled stranded and circular.

    Correct Answer
    D. Doubled stranded and circular.
    Explanation
    Bacteria's genetic material is double stranded and circular. This means that the DNA in bacteria consists of two strands that are twisted together in a double helix structure, and the ends of the DNA molecule are joined together to form a circular shape. This is in contrast to eukaryotic organisms, whose genetic material is double stranded and linear, with distinct ends. The circular nature of bacterial DNA allows for efficient replication and gene transfer, and it also provides stability to the genetic material.

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  • 42. 

    Derivatives of the toxin camptothecin are used to inhibit topoisomerase I. Which disease is linked to a defect in replication and cell division and might be effectively treated with these drugs?

    • A.

      Cancer

    • B.

      Infertility

    • C.

      Dwarfism

    • D.

      Cardiomyopathy

    Correct Answer
    A. Cancer
    Explanation
    Camptothecin derivatives are known to inhibit topoisomerase I, an enzyme involved in DNA replication and cell division. Cancer is a disease characterized by uncontrolled cell growth and division, often caused by mutations in genes involved in replication and cell division. Therefore, it is logical to assume that drugs inhibiting topoisomerase I, such as camptothecin derivatives, might be effective in treating cancer by targeting the defective replication and cell division processes.

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  • 43. 

    Which mutation would substitute TAA, TAG or TGA in place of an expressed codon?

    • A.

      Silent mutation

    • B.

      Missense mutation

    • C.

      Nonsense mutation

    • D.

      Frameshift mutation

    Correct Answer
    C. Nonsense mutation
    Explanation
    A nonsense mutation is a type of mutation that substitutes TAA, TAG, or TGA in place of an expressed codon. This mutation results in the premature termination of protein synthesis, leading to a nonfunctional or truncated protein. Silent mutations do not result in any change in the amino acid sequence, missense mutations substitute one amino acid for another, and frameshift mutations involve the insertion or deletion of nucleotides, leading to a shift in the reading frame and a completely different amino acid sequence.

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  • 44. 

    DNA polymerases can

    • A.

      Replicate 5׳ to 3.׳

    • B.

      Replicate 3׳ to 5.׳

    • C.

      Replicate at both the 3׳ and 5׳ ends.

    • D.

      None of the above.

    Correct Answer
    A. Replicate 5׳ to 3.׳
    Explanation
    DNA polymerases are enzymes responsible for synthesizing new DNA strands during replication. They can only add nucleotides in the 5' to 3' direction. This means that they can only replicate the DNA template strand in the 5' to 3' direction. Therefore, the correct answer is that DNA polymerases can replicate 5' to 3'.

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  • 45. 

    In bacteria, DNA replication will begin

    • A.

      At different origins and will be unidirectional.

    • B.

      At one origin and will be unidirectional.

    • C.

      At different origins and will be bidirectional.

    • D.

      At one origin and will be bidirectional.

    Correct Answer
    D. At one origin and will be bidirectional.
    Explanation
    In bacteria, DNA replication begins at one origin and is bidirectional. This means that replication starts at a specific point on the DNA molecule called the origin of replication, and proceeds in both directions away from this point. This allows for the efficient and simultaneous replication of both strands of the DNA molecule.

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  • 46. 

    If an enzyme degraded linker DNA, what would be an immediate consequence?

    • A.

      Nucleosomes would be unable to form

    • B.

      Nucleosomes would be unable to form a chromatin fiber

    • C.

      Chromatin fibers would not condense

    • D.

      DNA helix would be unable to form

    Correct Answer
    B. Nucleosomes would be unable to form a chromatin fiber
    Explanation
    If an enzyme degraded linker DNA, the immediate consequence would be that nucleosomes would be unable to form a chromatin fiber. Linker DNA is responsible for connecting nucleosomes, which are the basic units of chromatin. Without linker DNA, nucleosomes would not be able to organize into a higher order structure, resulting in the inability to form a chromatin fiber.

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  • 47. 

    Which type of mutation is likely to be the most disruptive to protein function?

    • A.

      Single base pair insertion

    • B.

      Three base pair deletion

    • C.

      Transition point mutation

    • D.

      Transversion point mutation

    Correct Answer
    A. Single base pair insertion
    Explanation
    Insertions or deletions cause frameshifts which typically lead to stop codons, creating a short, nonfunctional protein. A three base pair deletion, however, would maintain the reading frame and be less disruptive. Point mutations may be disruptive, but generally will be less so than a frameshift.

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  • 48. 

    Why do bacteria not need telomerase?

    • A.

      The lifespan of bacteria are short.

    • B.

      Bacteria cannot produce RNA.

    • C.

      Bacteria have a similar functioning enzyme.

    • D.

      Bacteria have circular DNA.

    Correct Answer
    D. Bacteria have circular DNA.
    Explanation
    Bacteria do not need telomerase because they have circular DNA. Unlike eukaryotic cells, which have linear DNA, bacteria have a circular chromosome. Circular DNA does not have ends that need to be protected and replicated by telomerase. Therefore, bacteria can continuously replicate their DNA without the need for telomerase, which is an enzyme that adds repetitive sequences to the ends of linear chromosomes to prevent them from shortening during replication.

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  • 49. 

    This characteristic of DNA allows the molecule to be a template which specifies the sequence of nucleotides in the opposite strand.

    • A.

      The strands are antiparallel

    • B.

      Complementary base pairing

    • C.

      DNA is a double helix

    • D.

      The backbone is phosphate and sugars

    Correct Answer
    B. Complementary base pairing
    Explanation
    Complementary base pairing refers to the specific pairing of nucleotide bases in DNA. Adenine (A) always pairs with thymine (T), and guanine (G) always pairs with cytosine (C). This complementary base pairing allows for the replication and transcription of DNA, as each strand can serve as a template for the synthesis of a new strand with a complementary sequence. The concept of complementary base pairing is essential for maintaining the genetic information encoded in DNA and ensuring the accurate transmission of this information during cell division and protein synthesis.

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  • 50. 

    The ends of linear eukaryotic chromosomes are maintained by the enzyme:

    • A.

      Primase.

    • B.

      Helicase.

    • C.

      Telomerase.

    • D.

      Topisomerase.

    Correct Answer
    C. Telomerase.
    Explanation
    Telomerase is the correct answer because it is the enzyme responsible for maintaining the ends of linear eukaryotic chromosomes. Telomeres are repetitive DNA sequences located at the ends of chromosomes that protect them from degradation and fusion with other chromosomes. Telomerase adds repetitive DNA sequences to the ends of chromosomes, counteracting the gradual loss of DNA during replication. This helps to preserve the integrity and stability of the chromosomes. Primase is involved in DNA replication, helicase unwinds the DNA double helix, and topoisomerase relieves the tension caused by unwinding. However, none of these enzymes specifically maintain the ends of chromosomes like telomerase does.

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