Ch. 3 HW ans key

# 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) 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, when the p value is < 0.05, null hypothesis is rejected. 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 heterozygotes. 9) Complete dominance, independent assortment, no gene interaction 10) The likelihood of rejecting the null hypothesis increases.

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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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## This note was uploaded on 02/29/2012 for the course BIO 325 taught by Professor Saxena during the Spring '08 term at University of Texas.

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

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