1 BIOLOGY 321 SPRING 2013 ANSWERS TO ASSIGNMENT SET #2 Answers to text questions: Chapter 2 http://fire.biol.wwu.edu/trent/trent/IGA_10e_SM_Chapter_02.pdf Chapter 3 http://fire.biol.wwu.edu/trent/trent/IGA_10e_SM_Chapter_03.pdf Answers to Edition 9 text questions: http://fire.biol.wwu.edu/trent/trent/answers2.9.pdf Answers to Questions from old quizzes and exams  Problem 1A (i). a (ii) c (iii) a (iv) d  Problem 1B a. aabb X AB  AaBb females showing dominant phenotypes and ab males showing recessive phenotypes b. AABB X ab  AaBb females and AB males – all showing dominant phenotypes c. AaBb X AB  ¼ females AABB ¼ AaBB ¼ AaBb ¼ AABb All females show dominant phenotypes and genotypes as above Males: ¼ AB ¼ Ab ¼ aB ¼ ab (phenotypes directly correspond to genotypes)2  Problem 2 The allele notation in a is correct since the a single-eye spot is the wild-type phenotype. Note that this allele designation conveys no information about dominance. The allele notation in e is correct as well since the cross shows a sex-linked inheritance pattern and the table (page 1) tells you that butterflies have Z and W chromosomes rather than X and Y. The allele notation in c is also correct since the outcome of the cross tells you that the wild-type allele is dominant to the mutant allele: Note ZZ male parent must be homozygous because only one phenotypic class of ZW female progeny were produced. double male ZeZe X single female ZEW  ZeW double females ZEZe single males  Problem 3 • Red and jagged are dominant over green and smooth. • If XY system, the wing trait would be autosomal - indicated by the fact that the phenotypes of the F1 males and females are the same for this trait in both crosses. The eye trait would be X-linked and in experiment #1 F1 males should have been green and jagged and F1 females red jagged. In experiment #2, F1 males and females should have been red and jagged. So XY system isn’t consistent with the data. • The data fit with the ZW (female) ZZ (male) system. • Wing trait is autosomal and eye trait is Z-linked. Designation of allele symbols: ZR Red eyes allele, Z linked dominant Zr Green eye allele, Z linked recessive J Jagged wing allele, autosomal dominant jSmooth wing allele, autosomal recessive answer continues on the next page3 Experiment 1 Parental: green, jagged female (ZrW, JJ) X red, smooth male (ZRZR, jj) F1: All progeny are red, jagged All female F1 progeny have the genotype (ZRW, Jj) All male F1 progeny have the genotype (ZRZr, Jj) F2: Females: 3/8 red, jagged (ZRW, J--) 3/8 green, jagged (ZrW, J--) 1/8 red, smooth (ZRW, jj) 1/8 green smooth (ZrW, jj) Males: 3/4 red, Jagged (ZR --, J--) 1/4 red, smooth (ZR --, jj) Males have 1/2 probability of being ZRZR and 1/2 probability of being ZRZr. Also, both male and female individuals in this generation displaying the dominant phenotype (J— ) have 1/3 probability of being JJ and 2/3 probability of being Jj. Experiment 2: Parental: red, smooth female (ZRW, jj) X green, jagged male (ZrZr,JJ) F1: all green, jagged females (ZrW, Jj) all red, jagged males (ZRZr, Jj) F2: Females: 3/8 red, jagged (ZRW, J--) 3/8 green, jagged (ZrW,J--) 1/8 red, smooth (ZRW, jj) 1/8 green, smooth (ZrW, jj) Males: 3/8 red, jagged (ZRZr,J--) 3/8 green, jagged (ZrZr,J-) 1/8 red, smooth (ZRZr,jj) 1/8 green, smooth (ZrZr,jj) All progeny in this F2 generation displaying dominant wing phenotype (J--) have 1/3 probability of being JJ and 2/3 probability of being Jj.4  Problem 4 Note the red flag in this problem: you don’t know anything about the ancestry of these flies, so you cannot assume that they are true-breeding. Also, you must inspect both crosses to determine dominance for the wing trait and whether the traits are autosomal or sex-linked. a. Round is dominant (from F1). Wingless is dominant: The 1:1 ratio in the F1 doesn’t tell you anything about dominance for the wing gene. But it does reveal that one parent must have been heterozygous for this trait. This means that the F1’s carrying the dominant allele must be het for the wing trait. Since the winged F1’s breed true, they must be homozygous and the wingless F1s heterozygous. b. Wing gene is autosomal since there is no difference in the segregation pattern in males and females in either cross. c. The eye gene is X linked because in the F2 progeny the phenotypic outcome for this trait is different for males and females. d. I’ll set up the parental genotypes and you can crunch through the F1 and F2 yourself. Parental w+w+; e+e+ X w+w; e Y e. w+w ; e+e F1 female X w+w+; e Y F2 male females: 1/4 winged and round w+w+; e+e 1/4 wingless and round w+w; e+e 1/4 wingless and oval w+w; ee 1/4 winged and oval w+w+; ee males: 1/4 winged and round w+w+; e+Y 1/4 wingless and round w+w; e+Y 1/4 wingless and oval w+w; e Y 1/4 winged and oval w+w+; e Y5  Problem 5 a. Green is dominant (from cross 1). b. The reciprocal crosses are consistent with XX/XY inheritance patterns (see below), but not ZZ,ZW inheritance patterns. Therefore, in Coolits, males carry the heteromorphic (XY) chromosome pair. c. ii (Mendel style) and v are correct. You could also use the Drosophila style of allele designation where the gene is named after the mutant phenotype and the wild-type allele is designated by the + superscript. So this is the green gene: g+ = the wildtype yellow allele g = mutant green allele Cross 1 Xg Xg green female crossed with a yellow Xg+ Ymale  all green progeny Xg+ Xg and Xg Y I’ll let you crunch through the genotypes for the second cross  Problem 6 a. 1/3 X 1/3 = 1/9 b. Two ways to work this problem: Strategy 1: Directly calculate chance that the one gene is het and the other homozygous dominant. Either AABb or AaBB meets the condition of the question, so you will sum the individual probabilities = (1/3)(2/3) + (2/3)(1/3) = 4/9 Strategy 2: Figure out probability of both homozygous (AABB) or both heterozygous (AaBb) and subtract from 1: Probability that one gene is het and the other homozygous dominant equals the probability that the plant is neither homozygous (AABB) nor heterozygous for both genes (AaBb) = 1.0 – [1/9 + 4/9] = 4/96  Problem 7 You discover a new