<p>BIO 208 answers to worksheet problems Unit 2</p><p>1. E, F, B, A, C, D, H, G 2. a. codominance b. expressivitiy c. epistasis d. penetrance e. random segregation f. incomplete dominance g. independent assortment h. allele i. dominance 3. pedigree 1 is not Y liked because the grandfather does not show the trait. It could be X-linked recessive if both mothers in generation I and II are heterozygote, XCXc. It could be autosomal recessive and cannot be autosomal dominant because a generation I individual would have to possess the trait. The second pedigree can be X-linked dominant with an XCXc mother, XCY son, OR autosomal dominant with a Cc mother, cc dad and Cc son, OR autosomal recessive with a cc mother, Cc dad and cc son 4. 1/256 5. BbWw X BbWw = 9/16 with B-W- = white 3/16are B-ww = black 3/16 B-ww = white 1/16 bbww = chestnut. This is a 12:3:1 ratio of phenotypes 6. a. XX male b. XY female c. AED d. hemophilia e. polydactyly f. PKU g. achondroplasia h. Down syndrome, trisomy 21, i. Patau j. Cri du chat k. alkaptonuria l. only these can be detected by karyotype: Down syndrome, Patau, Cri du Chat, XX male and XY female (only after birth) all the other disorders result from a single gene and would not be visible on a karyotype. 7. Y chromosome/SRY/TDF / testes/ testosterone /male embryo 8. XYfly is 1X/2A = 0.5 ratio = male XXY fly is 2X/2A = 1 = female XX fly with3 sets of autosomes is 2X/3A = 0.66 = intersex 9. ZSW female X ZsZs male = all silver males ZSZs and all gold females ZsW 10. a. Barr, b. Down, c. Edwards, d. Mendel, e. Sutton, f. Klinefelter 11. a. 46 b. 23 c. 45 d. 47 e. 69 f. 45 g. 46 h. 23 12. a. locus b. barr body c. trisomy d. translocation e. monosomy f. tetraploid 13. a. deletion b. inversion c. deletion d. mosaicism e. del, t, inv, dup, mos f. translocation g. deletion h. translocation, inversion i. translocation (if one chromosome attaches to another) j. duplication k. mosaicism 14. a. Phenylketonuria, b. phenylpyruvate toxic to brain cells, c. melanin pigment, d. alkaptonuria</p><p>15. a. F1, F2 b. Yy X Yy = 1/4 green in the F2 c. yy d. Tt X Tt = 3:1 ratio of T- (tall) to tt (short) e. incomplete dominance f. IAIA X IBIB gives all AB IAIO X IBIB gives 1/2 type AB and 1/2 type O IAIO X IBIO gives 1/4 AB, 1/4 A, 1/4 B, 1/4 O IAIA X IBIO gives 1/2 AB and 1/2 A g. 1/4 SY 1/4 Sy, 1/4 sY and 1/4 sy gametes from the plant h. SsYy X Ss Yy, a typical dihybrid cross i. 1/4 X/ 1/4 = 1/16 j. 1/2 X 1/2 = 1/4 k. compute these individual probabilities and add them: p(SSYY), p(SSyy), p(ssyy) and p(ssYY). This is 1/16 + 1/16 + 1/16 + 1/16 = 4/16 l. .XCXC X XcY = 1/2 carrier females and all normal males m. 1/2 affected males n. 0% o. 1:2:1 incomplete dominance, CBCW X CBCB = ½ black and ½ gray p. 50% q. Most likely X-linked recessive</p><p>Try additional problems in the back of the chapters, many answers are in the back of the book</p>