Problem Set 2_text_problems_key

Problem Set 2_text_problems_key - Biology 202 Problem Set 2...

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Biology 202 Problem Set 2 1/28/11 This problem set covers material from January 19 th through January 26 th From Griffiths et al.: Chapter 2, problems 4, 8, 20, 22, 24, 28, 29, 30, 38, 39, 46, 47, 48,52, 53, 64 Chapter 6, problems 7, 8, 9, 12, 13, 14, 16, 23 2.4) Because the DNA levels vary four-fold, the range covers cells that are haploid (gametes) to cells that are dividing (after DNA has replicated but prior to cell division). The following cells would fit the DNA measurements: x haploid cells 2x diploid cells in G1 or cells after meiosis I but prior to meiosis II 4x diploid cells after S but prior to cell division 2.8) In large part, this question is asking, why sex? Parthenogenesis (the ability to reproduce without fertilization – in essence, cloning) is not common among multicellular organisms. Parthenogenesis occurs in some species of lizards and fish, and several kinds of insects but it is the only means of reproduction for only a few of these species. In plants, about 400 species can reproduce asexually by a process called apomixis. These plants produce seeds without fertilization. However, the majority of plants and animals reproduce sexually. Sexual reproduction produces a wide variety of different offspring by forming new combinations of traits inherited from both the father and the mother. Despite the numerical advantages of asexual reproduction, most multicellular species that have adopted it as their only method of reproducing have become extinct. However, there is no agreed upon explanation of why the loss of sexual reproduction usually leads to early extinction or conversely, why sexual reproduction is associated with evolutionary success. On the other hand, the immediate effects of such a scenario are obvious. All offspring will be genetically identical to their mothers, and males would be extinct within one generation. 2.20) Do a testcross (cross to a/a). If the fly was A/A, all the progeny will be phenotypically A; if the fly was A/a, half the progeny will be A and half will be a. 2.24 a) You do not expect the mutation to be recessive. This would be an example of a haploinsufficient gene since one copy of the wild-type allele does produce enough protein product for normal function. b) An important assumption would be that having five of eight units of protein product would result in an observable phenotype. It also assumes that the regulation of the single wild-type allele is not affected. Finally, if the mutant allele was leaky rather than null, there might be sufficient protein function when heterozygous with a wild-type allele.
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2.28) Each die has six sides, so the probability of any one side (number) is 1/6. To get specific red, green, and blue numbers involves “and” statements that are independent. So each independent probability is multiplied together. a) (1/6)(1/6)(1/6) = (1/6)
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This note was uploaded on 02/07/2011 for the course BIOL 202 taught by Professor Smith during the Spring '11 term at Jefferson Davis Community College.

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Problem Set 2_text_problems_key - Biology 202 Problem Set 2...

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