Chapter 11 Test Questions

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Chapter 11 Test Questions
2015-11-29 18:12:14
Test Three: Zuzga
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  1. Explain the human nuclear genome.
    Approximately 3200 Mb of DNA, divided into 24 molecules, which consist of 22 autosomes and two sex chromosomes.

    Diploid somatic cells, which are all cells except the gametes, which are haploid

    46 chromosomes in all
  2. What is chromatin?
    Chromatin is the complex association between the DNA of the chromosomes and the proteins to which it binds, the latter including those responsible for packaging the DNA in a regular fashion within the chromosomes. It is half protein and half DNA. Histones are a group of histones, which contain basic amino acids and have little variation among different organisms.
  3. Explain the packaging mechanism of DNA
    Histones and DNA come together to form the nucleosome, which is the basic structural component of chromatin.

    The next level of chromatin structure is the 30-nm chromatin fiber, which reduces the length of the beads on a string by about 7 times.

    After the 30-nm fiber and euchromatin, there is the metaphase chromosome, which is the most compact form of chromatin in eukaryotes. Its formation may be due to the 30-nm fibers interacting with one another to draw the euchromatin structure into more compact conformations.
  4. What was the experiment that discovered nucleosomes?
    Biochemical work included nuclease protection experiments in which a DNA-protein complex was treated with an endonuclease, which cuts polynucleotides at internal phosphodiester bonds. The endonuclease has to gain access to the DNA in order to cut it and can only attach the phosphodiester bonds that are not masked by attachment to a protein. The sizes of the resulting DNA fragments were not random. Later EM observations of chromatin showed it to be beads on a string, which are the protein complexes attached to a DNA molecule.
  5. Explain the nucleosome.
    It has a core octamer consisting of two molecules each of H2A, H2B, H3, and H4, with the DNA molecule wound twice around each nucleosome. The only parts of DNA molecule not wound around the nucleosome is the linker DNA that join the individual nucleosomes to one another.

    Linker histones also exist, which are attached to each nucleosome to form the chromatosome, where it may act as a clamp, preventing DNA from detaching.
  6. Explain euchromatin.
    It is DNA that is not too condensed and still accessible for transcription of proteins. There are loops of DNA within the region attached to the nuclear matrix by AT-rich segments of DNA called matrix-associated regions (MARs) or scaffold attachment regions (SARs).
  7. Exlain heterochromatin.
    There is constitutive, which is a permanent feature of all cells and represents DNA that contains no genes and so can always be retained in a compact organization.

    There is also facultative, which is not a permanent feature but is seen in some cells some of the time and is thought to contain genes that are inactive in some cells or at some periods of the cell’s life cycle. When these genes are inactive, their DNA regions are compacted into heterochromatin.
  8. Explain the metaphase chromosome.
    It has two daughter chromosomes linked at their centromeres, the position where microtububles attach. The regions on either side are chromatids, or chromosome arms. Centromeric DNA is made up of repeat sequences called alphoid DNA, as well as a few genes.

    Nucleosomes containing the protein CENP-A instead of H3 may also bepresent. They are more compact and rigid. They are on the ouer surface and form an outer shell which kinetochores are constructured.
  9. Explain telomeres.
    They function to: (1) protect the ends of the DNA molecule curtained within the chromosome from attack by exonuclease enzymes; (2) protect the natural ends of the chromosomal DNA from DNA repair systems, which would join two chromosomes together if they sense damage; (3) prevent shortening of the strand

    Telomeric DNA is made up of hundreds of copies of a short sequence, where one strand is rich in G and the other is rich in C. The G-rich strand extends for up to 200 nucleotides beyond the terminus of the C-rich strand, giving a single-stranded overhang.

    TRF1 and TRF2 bind to these sequences, the former which regulates the length of the telomere and the latter which maintains the single strand extension.
  10. Why does the DNA strand get progressively smaller after every cell division?
    DNA synthesis cannot occur in the 3’ to 5’ position. Therefore, the lgging strand is synthesized in a series of Okazaki fragments, posing two problems. The first is tat the extreme 3’ end might not be copied because the final Okazaki fragment cannot be primed, resulting in a shorter strand. The second is if the last Okazaki fragment is placed at the extreme 3’ end of the lagging strand. Shortening occurs to a lesser extent because the terminal RNA primer cannot be converted into DNA because the methods for primer removal require extension of the 3’ end of an adjacent Okazaki fragment.
  11. How do telomeres solve the problem of shortening DNA?
    Though they can be copied in the normal fashion during DNA replication, they can also be extended by an independent mechanism catalyzed by the enzyme telomerase, which has both RNA and protein. Telomerase RNA has the reverse complement of the human telomere repeat and so can base-pair to the single strand DNA overhang, allowing extension, translocation of telomerase, and then extension again. Telomerase only adds to the end of the G-rich strand. However, C-rich strand elongation occurs when the G-rich strand is long enough. The primase-DNA polymerase alpha complex ataaches to the end of the C-rich strand and initiates synthesis of a new Okazaki fragment. It will still be shorter than the G-rich strand, bt the overall length of chromosomal DNA has not been reduced.
  12. What are the four phases of the cell cycle?
    • G1: longest phase; the cell grows, metabolizes, and performs its specialized function; the cell has its standard complement of chromosomes, 46 for a diploid human cell (may last for weeks or even months)
    • S: chromosomal DNA molecules are replicatied; replication origins are activated and bidirectional DNA synthesis progresses (6-8 hours)
    • G2: cell is tetraploid; at the end of G2, the chromosomes begin to condense to form the metaphase structures (3-4 hours)
    • M phase: nuclear and cell division take place
  13. How is DNA replication coordinated with the rest of the cell cycle?
    Coordination involves a variety of regulatory proteins, called cyclins. Some are active at the beginning of S and switch on DNA replication. Others prevent a second round of DNA replication during G2. They regulate events at the replication origin.
  14. Explain initiation of DNA replication at the origin.
    Initiation of DNA replication at an origin requires construction of a pre-replication complex (pre-RC), which is converted to a post-replication compex when the replication fork moves away from the origin. The post-RC can’t initiate replication, preventing segments of the genome from being recopied.

    The key protein involved in construction of a pre-RC is Cdc6p, which is syntheized during G1 and recruits the MCM helicase protein into the pre-RC, which breaks the base pairs iwthin the origin. A second group of pre-RC proteins are the replication licensing factors, which become bound to the chromosomal DNA toward the end of M phase and remain there until S phase, whre they are removed from the DNA. Their removal may be the key event that converts a pre-RC into a post-RC
  15. What are some control points of the cell cycle?
    The G1 to S phase transition is the major control point for genome replication.

    Initiation of replication doesn’t occur at the same time at all replication origins, nor is “origin firing” random. Some parts are replicated early in S phase and some later. The general pattern is that actively transcribed genes and the centromere are replicated early in S phase, and telomeres and nontranscribed regions later on. Early-firing origins are tissue-specific and reflect the pattern of gene expression.

    The positional effect may be a factor (as opposed to the sequence of the origin) in that it may be linked with the way that the DNA is packaged in different parts of the chromosome.

    The position of the origin in the nuclease may also be important as origins that become active at similar periods within S phase appear to be clustered together.

    DNA replication can also be brought to a halt during S phase if DNA damage is detected. With DNA damage comes activation of genes whose products are involved in DNA repair. The replication process can be arrested or slowed down.

    The cell can also be shunted into the pathway of apoptosis to prevent accumulation of DNA damage and the rise of a tumor.
  16. How is the appropriate partitioning of chromosomes assured?
    The answer lies in the structure of the metaphase chromosome, specifically the positioning of the kinetochores on the surface of the conjoined centromeres. Two kinetochores form, one on each centromere. When the microtubules attach to the chromosome, they create a tension that confirms that the correct attachments have been made.