Chapter 14 Test Questions

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Chapter 14 Test Questions
2015-11-29 19:18:38
Test Three: Zuzga
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  1. Explain meiosis I and meiosis II.
    Meiosis I results in two cells, each containing one copy of each of the replicated homologous chromosomes. The cells resulting from meiosis I are therefore diploid. During meiosis II, the attachment that holds each pair of daughter chromosomes together breaks.

    One daughter chromosome passes into one gamete, and the second daughter into the second gamete. The gametes are therefore haploid.
  2. Where are the differences between meiosis and mitosis seen?
    The first is when the replicated chromosomes condense and migrate to the middle of the nuclear region. During mitosis, homologous chromosomes remain separate from one another. In meiosis I, the pair up and form a bivalent.
  3. What are the two reasons for the variability between gametes?
    • The first results from the separation of the bivalents during anaphase I. The two homologous chromosomes have the same sets of genes but different alleles. When they get pulled apart, they randomly assort.
    • The second is recombination, when chromatids exchange segments of DNA. They can participate in different exchanges, producing different combinations of alleles.
  4. What is recombination involved in?
    It is involved in the breakage and reunion of DNA molecules, such as in the repair of damaged DNA molecules. It is also involved in the integration of bacterial genomes of DNA acquired by transformation, conjugation, or transduction, as well as the integration of episomes and lysogenic bacteriophage genomes.
  5. Explain homologous recombination.
    It begins when the two double-stranded molecules line up next to one another. A double-strand cut is made in one of the molecules, breaking this one into two pieces. One strand in each half is then shortened by removal of a few nucleotides, giving a 3’ overhang. The partnership is set up when one overhang invades the uncut DNA molecule, displacing one of its strands and forming a D loop. Strand invasion is stabilized by base pairing between the transferred segment of polynucleotide and the intact polynucleotide of the recipient molecule. The base pairing is possible because of the sequence similarity between the two molecules. The invading strand is extended by new DNA synthesis, enlarging the D loop, and at the same time, the other broken strand is also extended.

    After strand extension, the free polynucleotide ends are joined together. This gives a structure called a heteroduplex, in which the two double-stranded molecules are liked together by a pair of Holliday structures, which can move along the heteroduplex, called branch migration.
  6. Explain separation of the heteroduplex.
    Separation occurs by cleavage of the Holliday structure. There are two possible cleavage orientations.

    The first is a left to right cut across the chi form. This gives two molecules that have exchanged short segments of polynucleotides, with minor results, since there are similar sequences.

    The second is the up-down orientation, which results in reciprocal strand exchange, double stranded DNA being transferred between the two molecules so that the end of one molecule is exchanged for the end of the other molecule.
  7. Explain the RecBCD recombination system.
    The central player is the RecBCD enzyme, which is made up of three different proteins. Two—RecB and RecD –are helicases. To initiate homologous recombination, one copy of the RecBCD enzyme attaches to the free ends of a chromosome at a doble-starnd break. Using its helicase activity, RecBCD then progresses along the DNA until it reaches the first copy of a consensus sequence, called the chi site. The RecBCD then produces a double-stranded molecule with a 3’ overhang. The heteroduplex is then formed, which is mediated by RecA, whch forms the protein-coated DNA filament that is able to invade the intact double helix and set up the D loop. An intermediate in formation of the D loop is the triplex structure, a three-stranded DNA helix in which the invading polynucleotide lies within the major groove of the intact helix and forms hydrogen bonds with the base pairs it encounters. Once the heteroduplex has been established, the subsequent events are common to all three recombination systems.

    RuvA and RuvB catalyze the branch migration. Four copies of RuvA bind to the branch point, forming a core to which two RuvB rings, attach, one on either side. The resulting structure acts as a molecular motor, rotating the helices in the required manner. When branch migration ends, RuvAB complex detaches and is replaced by two RuvC proteins, which cleaves the Holliday structure.
  8. What is one puzzling aspect of homologous recombination?
    That is how they are resolved. Initially, it was thought that nucleases were responsible for that. However, they produce cleaed ends that cannot be ligated easily. It was eventually realized that a combination of helicases and topoisomerases can separate the heteroduplex back into independent chromoomes. GEN1 in human cells is similar to RuvC.

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