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Ch.3 HW ans key - Problems 1 A If G = gray and g = white...

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Ch. 3 HW answer key MCQs: 1) B 2) C 3) A 4) B 5) D 6) A 7) B T/ F: 1) T 2) F 3) F 4) T 5) T Short questions: 1) 9:3:3:1 2) Segregation is the separation of alleles during meiosis, while independent assortment states that a member of one gene pair has an equal and independent opportunity of segregating with either member of another gene pair. 3) Usually, when the probability value is less than 0.05 4) 3:1 5) This occurs in a cross involving doubly heterozygous individuals crossed to fully recessive individuals. The genes involved assort independently of each other. 6) No 7) ¼ X ½ = 1/8 8) No. In mating involving heterozygotes, three genotypic classes are expected in the offspring: fully dominant, fully recessive, and heterozygous. 9) Complete dominance, independent assortment, no gene interaction 10) The likelihood of rejecting the null hypothesis increases. Problems:
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Unformatted text preview: Problems: 1) A) If G = gray and g = white, then the probable genotype of the parents are Gg and Gg B) Genotypic = 1:2:1, phenotypic = 3:1 2) Phenotypes: wild, vestigial, hairy, vestigial hairy Numbers expected: { 9:3:3:1}: wild (576), vestigial (192), hairy (192), vestigial hairy (64) 3) a- (½) 7 = 1/128 b- (1/2) 7 chance of being all boys and (1/2) 7 chance of being all girls. 1/128+ 1/128 = 1/64 c- 7/128 d- 35/128 e- 35/128 4) a- ½ Aa X ½ Bb X ½ Cc X ½ Dd X ½ Ee = 1/32 b- ½ Aa X ½ bb X ½ Cc X ½ dd X ¼ ee = 1/64 c- ¼ aa X ½ bb X ¼ cc X ½ dd X ¼ ee = 1/ 256 d- No offspring with this genotype. The AaBbCcddEe parent can not provide a D and the other parent can not contribute a B allele. 5) a- Sally (Aa), mother (Aa), father (aa), and brother (aa). b- ½ c- ½...
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