Mutant 2 Mutant 3 A The offspring will be purple B The offspring will be white

Mutant 2 mutant 3 a the offspring will be purple b

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Mutant 2 Mutant 3 A) The offspring will be purple B) The offspring will be white a 2 a 2 BB AA b 1 b 1
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4. In plants, is being polyploid bad for the viability and health of the organism?A. Certainly – in most cases.B. Not necessarily – in many cases polyploid plants are viable.
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A. The human eyeB. Pocket watchesC. The human liverD. The human brain 1. The intelligent design proponent Michael Behe suggests that which is an example of ‘irreducible complexity’:
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A. Lord ByronB. James JoyceC. Alfred Russel WallaceD. Jane Goodall 2. The co-discoverer (along with Charles Darwin) of evolution through natural selection was:
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3. In which animal group can we see a broad spectrum of eye designs?
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4. Did this lecture change your views? A. No, I already supported the theory of evolution. B. No, I still support a biblical view of the origin of life. C. Yes, I feel more convinced about evolution now. D. Umm… I’m still thinking about it. Let me get back to you in a couple of years. Is it dinnertime yet?
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1. Are they a different breed or species of pigeon? A. Different breed B. Different species
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2. Are they a different breed or species of pigeon? A. Different breed B. Different species
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3. Are they a different breed or species of pigeon? A. Different breed B. Different species
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4. Are they a different breed or species of pigeon? A. Different breed B. Different species
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1. How many individuals does this gel represent?
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2. How many alleles does this gel represent?
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3. What would the genotype frequencies be? AA Aa aa 4 7 5 A. B. Freq. AA 0.25 0.20 Freq. Aa 0.44 0.65 Freq. aa 0.31 0.20 Answers
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The genotype frequencies give the relative number of each genotype in the population AA Aa aa 4 7 5 Frequency of AA = = 0.25 Frequency of Aa = = 0.44 Frequency of aa = = 0.31 4 16 7 16 5 16
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4. What would the allele frequencies be?AAAaaa475 p = A q = a A.p = 0.56 q = 0.44B.p = 0.53 q = 0.47C.p = 0.60 q = 0.40D.p = 0.47 q = 0.53
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The allele frequencies give the relative number of each allele in the population p =Frequency of A = AA Aa aa 4 7 5 = 0.47 q = Frequency of a = = 0.53 8 + 7 32 10 + 7 32 There are 16 individuals = 32 alleles
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5) Is this population at Hardy-Weinberg equilibrium? AA Aa aa 0.2 0.2 0.6 A) Yes B) No
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AA Aa aa 0.2 0.2 0.6 = 0.2 + ½ (0.2) = 0.3 p = Freq( A ) = 0.7 q = 1– p Expected frequencies, under Hardy-Weinberg: Freq(AA) = p 2 = (0.3) 2 = 0.09 Freq(Aa) = 2 pq = 2(0.3)(0.7) = 0.42 Freq(aa) = q 2 = (0.7) 2 = 0.49 Observed frequencies: Observed frequencies strongly depart from Hardy-Weinberg expectation.
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  • Spring '11
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  • Evolution, Zygosity, eye cancer, B. Analogy

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