380F11SolnPS3

# 380F11SolnPS3 - 01:447:380 Fall 2011 SOLUTIONS FOR PROBLEM...

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01:447:380 Fall 2011 SOLUTIONS FOR PROBLEM SET 3 Additional Problems 1. A) 1/64 = P(girl and affected) and (girl and affected) = ( ½ x ¼ ) x (1/2 x ¼ ); Multiply because of AND rule. B) 1/8 = P(girl and affected) = ½ x ¼ ; identical twins arise from a single fertilization event C) 81/256 = P(unaffected and unaffected and unaffected and unaffected) = ¾ x ¾ x ¾ x ¾ D) 175/256 = 1 – P(none of 4 children is affected) = 1 – 81/256 E) 9/256 = P(affected and affected and unaffected and unaffected) = ¼ x ¼ x ¾ x ¾ F) 9/64 = (3!/2!1!)(1/4) 2 (3/4) 1 2. X 2 = 0.50; d.f.=1; p>0.05, therefore, do not reject the hypothesis that sex ratio at birth is 1:1. X 2 =5; d.f.=1; p<0.05, therefore, reject the hypothesis that sex ratio at birth is 1:1. 3. A) P(aa and (BB or Bb) and (CC or Cc) and dd) = 3/128; B) 1 – [P(A_ and bb and C_ and D_) or (A_ and B_ and C_ and D_)] = 74/128; C) If you are A_B_ccD_, what is the probability that you are AaBbccDd? 2/3 x 1 x 1 x 2/3 = 4/9 4. A) 1; An allele (form of a gene) is present at only 1 specific location (locus) on a chromosome B) 2; There will be 1 locus on each of the 2 homologous chromosomes where any of these allele could reside. C) 2; It’s a diploid species, so 2 homologous chromosomes containing the loci for this allele will be present. D) 10 genotypic combinations in entire species; Let the alleles be A 1 , A 2 , A 3 , A 4 . Then possible genotypes include: A 1 A 1 , A 1 A 2 , A 1 A 3 , A 1 A 4 , A 2 A 2 , A 2 A 3 , A 2 A 4 , A 3 A 3 , A 3 A 4 , A 4 A 4 5. A) 2/3 curly : 1/3 normal B) CyCy homozygotes are lethal C) 90 curly and 90 normal Chapter 3 31. a) p(Aa and Bb and Cc and Dd and Ee) = ½ x ½ x ½ x ½ x ½ = 1/32 b) p(Aa and bb and Cc and dd and ee) = ½ x ½ x ½ x ½ x ¼ = 1/64 c) p(aa and bb and cc and dd and ee) = ¼ x ½ x ¼ x ½ x ¼ = 1/256 d) p(AA and BB and CC and DD and EE) = ¼ x 0 x ¼ x 0 x ¼ = 0

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35. a) The cross of two burnsi individuals produced both burnsi and pipiens offspring. The result suggests that these individuals were heterozygous with each possessing
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380F11SolnPS3 - 01:447:380 Fall 2011 SOLUTIONS FOR PROBLEM...

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