Problem Set 1_text_problems_key

# Problem Set 1_text_problems_key - Biology 202 Problem Set 1...

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Biology 202 Problem Set 1 1/21/11 This problem set covers material from January 10 th through January 24 th From Griffiths et al.: Chapter 1, problems 5, 6, 8, 9, 17, 19 Chapter 2, problems 2, 3, 9, 11, 13 Chapter 3, problems 7, 15, 16, 18, 20, 35, 36 Chapter 7, problems 1, 2, 3, 5, 18, 19, 24, 26, 27 1.5) If the DNA is double stranded, A=T and G=C and A+T+C+G = 100%. If T=15% then C = [100-15(2)]/2 = 35%. 1.6) If the DNA is double stranded, G=D=24% and A=T=26%. 1.8) Keeping this strand antiparallel to the strand given, the sequence would be: TAACCACGTAATGAAGTCCGAGA 1.9) Yes, There are no sequence restrictions in single-stranded DNA. The percentage of A must equal the percentage of T only in double stranded DNA. 1.17) a) For the fungus to be orange (as in mutant 1), orange pigment would have to accumulate and then not be converted into red pigment. This would occur if enzyme C was defective. b) Mutant 2 is yellow. Using similar logic as in part (a), enzyme B must be defective. c) An organism defective in both enzyme B and C would still have the phenotype associated with the block at the earlier step of this biochemical pathway. In this case, it would be due to lack of enzyme B. 1.19) Protein function can be destroyed by a mutation that causes the substitution of a single amino acid, even though the protein has the same immunological properties. For example, enzymes require very specific amino acids in exact positions within their active site. A substitution of one of these key amino acids might have no effect on overall size and shape of the protein while completely destroying its enzymatic activity. 2.2) PFGE separates DNA molecules by size. When DNA is carefully isolated from Neurospora (which has seven different chromosomes) seven bands should be produced using this technique. Similarly, the pea has seven different chromosomes and will produce seven bands (homologous chromosomes will co-migrate as a single band). 2.3) There is a total of 4 m of DNA and nine chromosomes per haploid set. On average, each is 4/9 m long. At metaphase, their average length is 13 μ m, so the average packing ratio is 13 x 10 - 6 m: 4.4 x10 -1 or roughly 1:34,000. This remarkable achievement is accomplished through the interaction of the DNA with proteins. At its most basic, eukaryotic DNA is associated with histones in units called nucleosomes and during mitosis, coils into a solenoid. As loops, it associates with and winds into a central core of nonhistone protein called the scaffold.

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2.9) As cells divide mitotically, each chromosome consists of identical sister chromatids that are separated to form genetically identical daughter cells. Although the second division of meiosis appears to be a similar process, the “sister” chromatids are likely to be different. Recombination during earlier meiotic stages has swapped regions of DNA between sister and nonsister chromosomes such that the two daughter cells of this division typically are not genetically identical.
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## This note was uploaded on 06/27/2011 for the course BIOL 202 taught by Professor Kieber-hogan during the Spring '08 term at UNC.

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Problem Set 1_text_problems_key - Biology 202 Problem Set 1...

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