Solutions to 100 passage-based Biology questions

Answer b is incorrect because the fitness of the

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Unformatted text preview: n is reduced fitness in a given population as a result of breeding of related individuals. Answer B is incorrect because the fitness of the heterozygous condition is not frequency dependent. Answer C is incorrect because genetic drift is a phenomenon that occurs in small populations which randomly lose genetic diversity. 72. B. Because the sickle cell trait affects the bloods ability to carry oxygen for cellular respiration, the production of CO2 and ATP in the electron transport chain are affected. Answer C is incorrect because fermentation is an anaerobic process and answer D is wrong because glycolysis does not involve oxygen. Answer A is also incorrect because oxygen is not required for the Krebs’ cycle. Oxygen is the final electron acceptor of the electron transport chain. 73. C. A woman living in the United States would not need protection from malaria (a rare disease in the US), nor wish to increase the chance her child would contract sickle cell anemia by reproducing with another carrier. Therefore, it would be most advantageous to be homozygous for the normal allele. 74. A. The substitution in beta hemoglobin is a missense mutation. A nonsense mutation would result in an inappropriate stop codon, shortening the resultant protein. A frameshift mutation is caused by an insertion or a deletion in the genetic code, and an RNA splicing mutation would affect transcript processing. 75. B. A deletion will cause a shift in the reading frame, changing a large portion of the resultant protein. A base change at the third position (the 3’ end) is less likely to change the matching amino acid than a change in the first position (5’ end). 76. C. This question involves basic understand of Hardy-Weinberg equilibrium. The frequency of the sickle cell trait is 0.12. Therefore the frequency of the normal trait is 0.88 (p + q = 1; 0.12 + 0.88 = 1). To determine the percentage of homozygous and heterozygous individuals, the previous equation is squared: p2 + 2pq + q2 = 1; p2 is the percentage of individuals homozygous for the normal trait, q2 is the percentage of ind...
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This note was uploaded on 03/23/2011 for the course CHEM 100 taught by Professor Vallis during the Spring '11 term at Dalhousie.

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