genetics exam 2 problem sets

genetics exam 2 problem sets - 1 Caitlin Favaloro CBF229...

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1 Caitlin Favaloro CBF229 T/Th 12:30pm class Problems and Questions for Chapter 6 Material: C2. Conjugation is not a form of sexual reproduction because it does not involve the mixing of two genomes to produce gametes. However, it is similar to reproduction because it involved the direct transfer of genetic material, and the genetic material in the new cell is somewhat mixed. C9. During the lytic cycle, a bacteriophage injects its DNA into the cytoplasm of a cell. The phage DNA then directs the synthesis of new phages within that cell; the cell bursts; and many new bacteriophages are released. In the lysogenic cycle, the phage integrates its DNA into the host’s chromosome creating a prophage. Prophages exist in the dormant state for a while, causing no new bacteriophages to be made and the cell remains unharmed. The prophage cell then replicates and divides mitotically to produce two daughter cells with the same prophage. Eventually the prophage can become activated and enter the lytic cycle. C13. In bacterial transformation, the donor cell “dies” and part of its fragmented material enters a competent recipient cell. The fragmented genetic material must break through the cell wall and cell membrane to enter into the new cell, and recombination must occur for the variation to become heritable. A competent cell is a cell that is able to become transformed C14. The transfer of conjugative plasmids (such as F factor DNA) does not require recombination. C15. The process would be much more efficient because the donor cell would not have to fragment its own DNA and insert part of it into a new cell; rather, its genetic material can remain in tact and variation will end up occurring in that cell instead. E1. Based on figure 6.1 the formation of the colonies is due to the conjugation of the two strains, not the rare double mutations E3. The U-tube apparatus can distinguish between transduction and conjugation, because it allows conjugation to occur and does not allow transduction to occur. This is because the filter is too small to allow the phage to pass and directly inject its DNA into the host bacteria. The U tube apparatus cannot distinguish between transformation and transduction because transformation would not be able to occur either. This is because the DNA would not be able to break through the recipient’s cell wall and cell membrane. E10. You can treat the P1 lysate with Dnase I. If the DNA is within a P1 phage, it will be protected from Dnase I. Therefore, one can distinguish between transformation (which would be inhibited by Dnase I) and transduction (which will not be inhibited by DnaseI).
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2 Additional Questions: A1. The abbreviation lac is for lactose, a sugar; leu is for leucine, an amino acid; and str is for streptomycin, an antibiotic. Consider each of the mutant strains below in comparison to the wild-type strain with genotype lac + leu + str S , and write S if the indicated
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genetics exam 2 problem sets - 1 Caitlin Favaloro CBF229...

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