VUBio110BSp'08-Exam3Key

VUBio110BSp'08-Exam3Key - EXAM #3 - KEY BSCI 110B Section 2...

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EXAM #3 - KEY BSCI 110B Section 2 – Intro to Biological Science – Spring 2008 26 March 2008 Denise Due-Goodwin Multiple Choice: 1. d 2. e 3. d 4. c 5. c 6. c 7. d 8. d 9. b 10. b 11. a 12. d 13. d 14. d 15. c 16. d 17. b 18. a 19. b 20. c 21. b 22. c 1. Yes, it is possible…A variety of scenarios or explanations would be acceptable. One example would be: parent genotypes Aabbcc x Aabbcc offspring could inherit “A” from each parent, AAbbcc, so the additive effect of this continuous trait would produce a darker-skinned child. *Any similar explanation that is accurate would be acceptable. Could also even use explanation of the contribution of environmental effects in association with the genotype (norm of reaction). A child could be genetically similar to parents, yet environment (“tanning”) could cause darker skin. 2. Frequency of carriers (i.e., heterozygotes) = 2pq from Hardy-Weinberg: 1/2500 = carriers = q 2 = 0.0004 q = 0.02 So, 2pq = 0.0392 p = 0.98
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Chance of two individuals from population both being carriers having normal child (3 pts.)Carrier frequency from above – product (multiplication) rule: (0.0392) (0.0392) (3 pts.) Chance of phenotypically normal child: 0.75 or ¾ ; product rule again (0.0392) (0.0392) (0.75) = 0.00115 3A. (1 pt. for each correct similarity or difference) Similarities (most likely responses)
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This note was uploaded on 04/07/2008 for the course BSCI 110B taught by Professor Due during the Spring '08 term at Vanderbilt.

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VUBio110BSp'08-Exam3Key - EXAM #3 - KEY BSCI 110B Section 2...

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