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Name____________________________________ Page 1 of 6 Short answer (5 questions @ 2 pts each) 1a) __AA × aa_(no partial credit)__ For the genotypes AA, Aa and aa, what cross yields all Aa progeny? b) ___Aa × Aa_(no partial credit)__ For the genotypes AA, Aa and aa, what cross yields AA, Aa and aa progeny? c) ___1:2:1 (no partial credit)_____ In Problem 2, what is the ratio of AA : Aa : aa progeny? d) __1x(1/2)x(1/2)x1x(1/2) = 1/8___ An individual of genotype AA Bb Cc DD Ee is (no partial credit) testcrossed. Assuming that the loci undergo independent assortment, what fraction of the progeny are expected to have the genotype Aa Bb Cc Dd Ee? e) ____anaphase (2 pts)__________ Stage of mitosis or meiosis in which anaphase I (1 pt) chromosomes oriented on opposite sides of the metaphase plate undergo separation. 2. (10 points) There is a strain of D. melanogaster that is pure-breeding for white eyes and another strain that is pure-breeding for red eyes. When white-eyed females are crossed to redeyed males, the progeny are red-eyed females and white-eyed males. a) Explain this result. Give the genotypes of the parents and progeny and define all symbols used. (10 pts) Xw = white eye allele on X chromosome is recessive X+ = red eye allele on X chromosome is dominant Y = Y chromosome P0: F1: XwXw female x Xw X+Y male X+ Y XwX+ red female XwY white male XwX+ XwY red female white male Initial description must include symbols (1 pt) and fact that white is recessive (1 pt). Correct genotypes of parents (2 pts). Correct genotypes (6 pts) of F1 (full credit if male shown with 1 X, but no Y). (-1 to -2) if genotypes shown, but not which male and which female. If postulated the reverse (female heterogametic, white dominant), full credit if adequately explained and internally consistent (would also have to postulate in part b that ND occured in male, which no one did). Xw Name____________________________________ Page 2 of 6 3. (20 points) Type 1 albinism in humans is a rare hereditary condition associated with a reduced amount of tryosinase, an enzyme required for the conversion of the amino acid tyrosine to the dark pigment melanin. In the following pedigree, males are indicated by squares, females by circles and individuals with type 1 albinism are indicated by shading. A B C D E a) What is the mode of inheritance of type 1 albinism? Explain your reasoning. (5 pts) Recessive (1 pt) because unaffected parents have affected progeny (1 pt). Autosomal (1 pt) since unaffected father has affected daughter (2 pt); affects both males and females (1 pt). b) List the genotypes of the indicated family members as specifically as possible with the data available. If there is more than one possibility, clearly indicate all possibilities. Except where necessary to explain the pedigree, assume that the individuals marrying into the family are homozygous wild-type. (5 pts) 1 pt each Individual A Aa or AA (or A-) Individual B Aa or AA (or A-) Individual C Aa Individual D Aa Individual E Aa or AA (or A-) c) If individuals B and E marry, what is the probability that their first child will be affected by type 1 albinism? Show your work. (5 pts) B’s father is Aa, mother is AA probability B is Aa = 1/2 (2pts) Name____________________________________ Page 3 of 6 E’s parents are BOTH Aa. E is NOT aa probability E is Aa = 2/3 (2pts) probability that a child of 2 carriers is affected = 1/4 (1 pt) Probability of B being carrier, E being carrier, and producing and affected child = (1/2)x(2/3)x(1/4) = 2/24 = 1/12 -1 if equation incorrect; -1 for math error. d) If their first child is affected, what is the probability that their second child will be unaffected? Show your work. (5 pts) Both B and E are carriers (2 pts); child has 3/4 chance of being unaffected (3 pts). Child has 1/4 chance of being affected (2 pts if specify 'affected child'; 1 pt if not). If missed the point that B and E are carriers, but correctly determined probability as if that were not known (1 - answer in c) (1 pt). 4. (24 points) Female Drosophila melanogaster heterozygous for three recessive alleles a, b, and c are crossed to abc males. The phenotypes of 10,000 progeny are scored as follows: +++ 4000 abc 4020 +bc 750 a++ 730 +b+ 230 a+c 250 ++c 8 ab+ 12 a) What results would you expect if a, b, and c were unlinked? (4 pts) All 8 phenotypic classes would appear at approximately equal frequency -or1:1:1:1:1:1:1:1 -or- 1250 of each class. b) Based on the observed results, what is the map order? Explain your reasoning. (4 pts) A-C-B -or- B-C-A (1 pt) Least common genotypes represent double-crossover classes (1 pt); most common represent parental classes (1 pt); gene that differs is in the middle (1 pt). Full credit if calculated map distances to determine. Name____________________________________ Page 4 of 6 c) Using all of the relevant data, calculate each of the three two-factor recombination frequencies. Show your work. Express all frequencies as percentages. (6 pts) Recombination between ac cb +++ abc +bc a++ +b+ a+c ++c ab+ 4000 4020 750 730 230 250 8 12 750 730 8 12 1500 230 250 8 12 500 ab 750 730 230 250 *double recombinant 2(8)* count 2X for outside 2(12)* genes 2000 (may also add values for ac and cb) recomb. freq. between a + c = Rac = 1500/10,000 = 15% (2 pts); if missed DCO's (1 pt) recomb. freq. between c + b = Rcb = 500/10,000 = 5% (2 pts); if missed DCO's (1 pt) recomb. freq. between a + b = Rab = 2000/10,000 = 20%(2 pts); if missed DCO's (1 pt) d) What is the frequency of double recombination? Show your work. (5 pts) Double recombinants are: ++c 8 ab+ 12 Frequency = (8+12)/10,000 = 20/10,000 = 0.2% If used 2 x DCO's and calculated 0.4% (2 pts); if calculated predicted rather than observed (0.75%, answer for next part) (1 pt). f) What frequency of double recombination would you calculate if the data contained no evidence of interference? Show your work (5 pts) Rac x Rcb = 15% x 5% = 0.75% Use of absolute number '75' (3 pts); definition of interference (1 pt). Name____________________________________ Page 5 of 6 Question 1 (24 pts) 6 pts A) Draw and label the leading and lagging strands on this diagram. Be sure to label the 5´ and 3´ ends of both strands. 2 pts B) Indicate where helicase would be found and its direction of movement. 4 pts C) Indicate the one or more sites where DNA polymerase would be found and its/their direction of movement. 8 pts D) i) ii) iii) iv) yes no no no yes yes no yes 4 pts E) Remember....the primer is RNA so 3´ UUAAGCAUAU 5´ 6a. If you place a culture of eukaryotic cells into medium containing 3H-thymine just prior to S, how will this radioactivity be distributed in a pair of homologous chromosomes at metaphase? Would the radioactivity be in (a) one chromatid of one homolog, (b) both chromatids of one homolog, (c) one chromatid each of both homologs, (d) both chromatids of both homologs, or (e) some other pattern? Choose the correct answer and explain your reasoning. (d) both chromatids of both homologs — replication is semi-conservative so one strand of each chromatid will be "new" and radioactive. (2 pts): d (-1 if right answer for wrong reason) (4 pts): explanation or diagram; for full credit must mention that one strand of each chromatid is radioactive. Partial credit (2 pts) if answered 'c' because confused chromatids and DNA strands, if provided consistent explanation. b. If these same cells are then allowed to go through a second S phase but with only nonradioactive nucleotides available, how would you expect the radioactivity to now be distributed in a pair of homologous chromosomes at metaphase? (Ignore the effect recombination might Name____________________________________ Page 6 of 6 have on this outcome.) Would the radioactivity be in (a) one chromatid of one homolog, (b) both chromatids of one homolog, (c) one chromatid each of both homologs, (d) both chromatids of both homologs, or (e) some other pattern? Choose the correct answer and explain your reasoning. (c) one chromatid of both homologs — after the first division, each chromosome has one strand that is radioactive and one that is not. When DNA replication occurs again, one chromatid gets this radioactive strand as its "old" strand while the other chromatid gets the non-radioactive strand as its "old" strand. The new strands are not radioactive so the result is (c). (2 pts): c (-1 if right answer for wrong reason) (4 pts): explanation or diagram Partial credit (1 pts) if answered 'e' or 'a' because confused chromatids and DNA strands, if provided consistent explanation