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Two Base Pair of DNA Molecules - Final Exam | BI 381, Exams of Molecular biology

Material Type: Exam; Professor: Gathman; Class: Molecular Genetics; Subject: Biology; University: Southeast Missouri State University; Term: Fall 2006;

Typology: Exams

Pre 2010

Uploaded on 08/08/2009

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Bi 381 Final Exam Fall 2006
1. Relax.
2. Write your name on the back of the last page ONLY.
1. (15 points) Fill in the rest of this two base-pair DNA molecule. Show covalent bonds
as lines, hydrogen bonds as dotted lines, double bonds as double lines, carbons in a
ring as corners, and other atoms as their atomic symbols. You need not show
hydrogens unless they are part of hydrogen bonds or have some other functional
significance.
pf3
pf4
pf5

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Bi 381 Final Exam Fall 2006

  1. Relax.
  2. Write your name on the back of the last page ONLY.
  3. (15 points) Fill in the rest of this two base-pair DNA molecule. Show covalent bonds as lines, hydrogen bonds as dotted lines, double bonds as double lines, carbons in a ring as corners, and other atoms as their atomic symbols. You need not show hydrogens unless they are part of hydrogen bonds or have some other functional significance.
  1. (10 points) Avery, MacLeod, and McCarty mixed heat-killed IIIS cell extract with chymotrypsin, a protease. They then mixed the treated IIIS cell extract with live IIR cells. They found that some IIR cells were transformed into live IIIS cells. What could they conclude about the transforming material based on this particular experiment? Explain.
  2. (5 points each) For each of the following types of lesions in E. coli , tell whether it would be most likely to be repaired by direct reversal, base excision repair, mismatch repair, or nucleotide excision repair, and justify your answer. a. spontaneous deamination of adenine results in the base hypoxanthine in the DNA. This is a very common event. b. Cosmic rays produce free radicals in the cell. As a result, a large portion of a triglyceride lipid molecule becomes bound to a cytosine in the DNA.
  1. (10 points) You have found an E. coli mutant in which β-galactosidase is never galactosidase is never produced in significant quantities. Propose one mutation affecting the positive control mechanism in the lac operon that could produce this effect. Tell what gene or site is mutant and how it produces the observed phenotype.
  2. a. (5 points) A baby, Carmen, is a mosaic of XXX and XO cells. How did this condition arise, in what kind of cell division, in whom, when,? b. (5 points) Diagram the abnormal cell division described in your answer to part a.
  1. (5 points each) a. Why are human trisomy-galactosidase is never 21 infants relatively common, while no trisomy-galactosidase is never 6 babies are observed? Explain. b. Avocados have N=12. You've produced a triploid avocado tree. What is the probability that a pollen grain from this tree will be euploid? Show your work. c. There are lots of genes on the X chromosome. Why is it that trisomy-galactosidase is never X humans have relatively few phenotypic abnormalities? Explain.