Sample Problems: Give genotypes of parents: 1. Round, yellow Round, yellow 303 Round, green 95 2. Round, yellow Round, yellow 94 Round, green 96 3. Round, yellow Round, yellow 213 Round, green 71 X wrinkled, yellow wrinkled, yellow 297 wrinkled, green 105 X wrinkled, green wrinkled, green 95 wrinkled, yellow 93 X Round, yellow wrinkled, yellow 70 wrinkled, green 24 Sample Problems: 1. Round, yellow Ww , Gg Round, yellow 303 (3) Round, green 95 (1) 2. Round, yellow Ww , Gg Round, yellow 94 (1) 213 (9) Round, green 71 (3) wrinkled, yellow ww , Gg wrinkled, yellow 297 (3) X Round, green 96 (1) 3. Round, yellow Ww , Gg Round, yellow X wrinkled, green ww , gg wrinkled, green 95 (1) X wrinkled, green 105 (1) wrinkled, yellow 93 (1) Round, yellow Ww , Gg wrinkled, yellow 70 (3) wrinkled, green 24 (1) Using the information given, fill in all blanks below. Circle each gamete. A dihybrid cross for two autosomal traits in the guinea pig for hair type. L = short, l = long (complete dominance); W'W' = yellow, WW' = cream, WW = white (semidominance). P Phenotypes Genotypes Gametes Short, yellow LLW'W' F1 Phenotypes Genotypes Gametes F2 F2 F2 F2 x x x Long, white llWW . . x x x . . . Genotypes Phenotypes Phenotypic ratio Genotypic ratio . . . . USE THE INFORMATION ABOVE TO MAKE A TESTCROSS BELOW. F1 Parent Phenotypes Genotypes Gametes Testcross Parent x x x . . . Testcross Progeny Phenotypes Genotypes . Testcross Phenotypic Ratio . Testcross Genotypic Ratio . 41 Using the information given, fill in all blanks below. Circle each gamete. A dihybrid cross for two autosomal traits in the guinea pig for hair type. L = short, l = long (complete dominance); W'W' = yellow, WW' = cream, WW = white (semidominance). P Phenotypes Genotypes Gametes Short, yellow LLW'W' 1/1 LW’ F1 Phenotypes Genotypes Gametes Short, cream x Short, cream LlW’W x LlW’W 1/4 LW’ 1/4 LW 1/4 lW’ 1/4 lW x 1/4 LW’ 1/4 LW 1/4 lW’ 1/4 lW F2 F2 F2 F2 x x x Long, white llWW 1/1 lW . . . . Genotypes 32 = 9, (1+2+1)2 Phenotypes (21)(31) = 6, (3+1)(1+2+1) Phenotypic ratio 6:3:3:2:1:1 Genotypic ratio 4:2:2:2:2:1:1:1:1 . . . . USE THE INFORMATION ABOVE TO MAKE A TESTCROSS BELOW. F1 Parent Phenotypes Short Genotypes Ll Gametes 1/2 L 1/2 l . Testcross Parent x Long x ll x 1/1 l Testcross Progeny Phenotypes Genotypes Short Ll Long ll Testcross Phenotypic Ratio 1:1 Testcross Genotypic Ratio 1:1 . . . . . . . 41 . FORKED LINE METHOD FOR SOLVING CROSSES P DDGGWW x ddggww F1 DdGgWw x DdGgWw F2 3 round = 27 tall, yellow, round 3 yellow 1 wrinkled = 9 tall, yellow, wrinkled 3 tall 3 round = 9 tall, green, round 1 green 1 wrinkled = 3 tall, green, wrinkled 3 round = 9 dwarf, yellow, round 3 yellow 1 wrinkled = 3 dwarf, yellow, wrinkled 1 dwarf 3 round = 3 dwarf, green, round 1 green 1 wrinkled = 1 dwarf, green, wrinkled 44 P DDGGWW x ddggww F1 DdGgWw x DdGgWw F2 GG DD 2 Gg gg GG 2 Dd 2 Gg gg Genotypic frequency Phenotypic ratio Phenotypes Genotypes Tall, Yellow round DDGGWW DDGGWw DDGgWW DDGgWw DdGGWW DdGGWw DdGgWW DdGgWw 1 2 2 4 2 4 4 8 27 DDGGww DDGgww DdGGww 1 2 2 9 2 WW = 1 DDGGWW Ww = 2 DDGGWw ww = 1 DDGGww 2 WW = 2 DDGgWW Ww = 4 DDGgWw ww = 2 DDGgww 2 WW = 1 DDggWW Ww = 2 DDggWw ww = 1 DDggww 2 WW = 2 DdGGWW Ww = 4 DdGGWw ww = 2 DdGGww Tall, yellow wrinkled DdGgww 2 WW = 4 DdGgWW Ww = 8 DdGgWw ww = 4 DdGgww Tall, green, round DDggWW DDggWw DdggWW DdggWw 2 WW = 2 DdggWW Ww = 4 DdggWw ww = 2 Ddggww Tall, green wrinkled DDggww Ddggww 45 4 1 2 2 4 9 1 2 3 GG dd 2 Gg gg 2 WW = 1 ddGGWW Ww = 2 ddGGWw ww = 1 ddGGww 2 WW = 2 ddGgWW Ww = 4 ddGgWw ww = 2 ddGgww 2 WW = 1 ddggWW Ww = 2 ddggWw ww = 1 ddggww Genotypic frequency 1 2 2 4 Phenotypic ratio 9 Phenotypes Dwarf, yellow, round Genotypes ddGGWW ddGGWw ddGgWW ddGgWw Dwarf, yellow wrinkled ddGGww ddGgww 1 2 3 Dwarf, green round ddggWW ddggWw 1 2 3 Dwarf, green wrinkled ddggww 1 1 (a) (b) (a) Cross between two F1 garden peas of the genotype DdGgWw. The forked-line method is employed and the genotypes are illustrated. These results represent the F2 of a cross similar to those obtained from the Punnett square method involving 64 squares. (b) Summary of F2 from trihybrid cross resulting in a 27:9:9:9:3:3:3:1 phenotypic ratio. 45 cont. METHOD FOR SOLVING TESTCROSS TYPE PROBLEMS INVOLVING THREE GENE PAIRS. THIS CROSS IS BETWEEN AN F1 GARDEN PEA WITH TALL VINES, YELLOW AND ROUND SEEDS, AND THE FULLY RECESSIVE PARENTAL TYPE WITH DWARF VINES, GEEEN AND WRINKLED SEEDS. F1 Parent Tall, yellow, round Seed parent DdGgWw x DGW DGw DgW Dgw dGW dGw dgW dgw Testcross Parent Dwarf, green, wrinkled Pollen parent ddggww dgw Gametes DGW DGw DgW Dgw dGW dGw dgW dgw dgw Phenotypes Tall, yellow, round Tall, yellow, wrinkled Tall, green, round Tall, green, wrinkled Dwarf, yellow, round Dwarf, yellow, wrinkled Dwarf, green, round Dwarf, green, wrinkled DGW DGw DgW Dgw dGW dGw dgW dgw dgw dgw dgw dgw dgw dgw dgw dgw Genotypic Genotypes frequency DdGgWw 1 DdGgww 1 DdggWw 1 Ddggww 1 ddGgWw 1 ddGgww 1 ddggWw 1 ddggww 1 1. A 1:1 phenotypic ratio from a testcross indicates a monohybrid. 2. A 1:1:1:1 phenotypic ratio from a testcross indicates a dihybrid. 3. A 1:1:1:1:1:1:1:1 phenotypic ratio from a testcross indicates a trihybrid. 46 Phenotypic ratio 1 1 1 1 1 1 1 1 P DD x dd D – Tall d - Dwarf F1 Dd Number of different heterozygous gene pairs in F1 genotype: n Number of different F1 gametes = 2n: D d Number of different F2 genotypes = 3n: DD, Dd, dd Number of different F2 phenotypes = 2n: Tall, dwarf Number of F2 progeny required to get all possible combinations in the correct proportions = 4n: DD Dd Dd dd Probability of getting one specific combination of alleles in a gamete = 1/2n Relations among pairs of independent alleles, gametes, F2 genotypes and F2 phenotypes when dominance is present: _______________________________________________________________________________ Number of Number of Number of Number of Number of One specific F2 F2 combination of heterozygous kinds of F2 progeny alleles in a gamete pairs F1 gametes genotypes phenotypes 1 2 3 2 4 1/2 2 4 9 4 16 1/4 3 8 27 8 64 1/8 4 16 81 16 256 1/16 10 1024 59049 1024 1,048,576 1/1024 n n n n 3 2 4 (1/2)n n 2 _______________________________________________________________________________ F2 phenotypic ratio (1 pair): 3:1 (2 pairs): (3:1)2 (3 pairs): (3:1)3 F2 genotypic ratio (1 pair): 1:2:1 (2 pairs): (1:2:1)2 (3 pairs): (1:2:1)3 48 List the number of ways the following exponents can be used: 2n 3n 4n (1/2)n EXAMPLE PROBLEM F1 Aa n No. diff. Gametes No. diff. F2 Geno. No. diff. F2 Pheno. No. F2 Progeny BbCc DdEeFf GgHhMmNn Example problem: P F1 AAbbCCddEE x aaBBccDDee n = _____ AaBbCcDdEe 49 No. diff. F1 Gametes No. diff. F2 Geno. No. diff. F2 Pheno. No. F2 Progeny ______ ______ ______ ______ EXAMPLE PROBLEM F1 Aa n 1 No. diff. Gametes 2n =2 No. diff. F2 Geno. 3n =3 No. diff. F2 Pheno. 2n = 2 No. F2 Progeny 4n = 4 BbCc 2 DdEeFf 3 GgHhMmNn 4 4 8 16 9 27 81 4 8 16 16 64 256 Example problem: P F1 AAbbCCddEE x aaBBccDDee n= AaBbCcDdEe 49 5 No. diff. F1 Gametes 32 No. diff. F2 Geno. 243 No. diff. F2 Pheno. 32 No. F2 Progeny 1,024 . . . . The following cross is made: AaBbccDDEe x AaBbCcddEE • 1) How many different phenotypes might you expect to get in the offspring? • 2) How many different genotypes might you expect to get in the offspring? AaBbccDDEe x AaBbCcddEE • 1) How many different phenotypes might you expect to get in the offspring? 22 x 2 x 1 x 1 = 8 • 2) How many different genotypes might you expect to get in the offspring? 32 x 2 x 1 x 2 = 36 Assuming independent assortment, what is the probability of the offspring of the cross AaBbccddEe x AabbccDdEe having a aabbccddee genotype? Answer: 1/4 x 1/2 x 1/1 x 1/2 x 1/4 = 1/64 Aa BB cc Dd EE × AA Bb Cc dd Ee • Number of gamete types produced by each parent? • Probability for progeny with Aa BB cc dd Ee genotype? • Probability for progeny with A_ B_ C_ D_ E_ phenotype? • Number of possible genotypes in progeny? • Number of possible phenotypes in progeny? • Genotypic ratio of progeny? • Phenotypic ratio of progeny? • Genotypic ratio of progeny? • Perform testcross on each parent separately and give testcross results phenotypically and genotypically- give testcross genotypic and phenotypic ratios. Aa BB cc Dd EE × AA Bb Cc dd Ee Number of possible genotypes in progeny = 32 (2) (2) (2) (2) (2) = (2)5 = 32 Genotypic ratio of progeny = (1 + 1) (1 + 1) (1 + 1) (1 + 1) (1 + 1) = (1 + 1) 5 = (1 + 1 + 1 + 1) (1 + 1 + 1 + 1) (1 + 1) = (1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1) (1 + 1) =1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1 +1+1+1+1+1+1+1+1+1+1+1+1+1 = 1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1 . . . . . . Testcross Progeny Genotypic/Phenotypic ratio of : 1:1:1:1:1:1:1:1:1:1:1:1:1:1:1:1 Give genotype of unknown dominant phenotype and TP (testcross parent) Number of F2 genotypes possible? Number of F2 phenotypes possible? Number of F1 gamete types? Common denominator of proportions for each genotypic and phenotypic category? Probability for one specific combination of alleles in one gamete type? Ll WW′ × Ll WW′ F2 results and ratios Phenotypic Genotypic Testcross results . Number of F gamete types? . Use 3 to predict number of F . Number of F phenotypes? 1 n 2 2 genotypes. Ll WW′ XX ZZ × Ll WW′ XX ′ ZZ ′ . Give Genotypic and Phenotypic results -ratios . Total number of genotypes and phenotypes in progeny? . Perform testcross using each parent separatelygive testcross genotypic/phenotypic ratios. A = Red aa = White B = Long bb = Short Red, Long × Red, Short A_B_ × A_ bb 3 : 1 : Red, White, Long Long Genotypes of Parents? Answer: AaBb × Aabb 3 Red, Short : 1 White, Short Gene Interaction • Although genes may affect phenotypes differently, the expression of a gene will sometimes modify the expression of another non-allelic gene. GENE INTERACTION A B C D Two different allelic pairs interact in affecting comb shape in the fowl. A. rose (R-pp) B. pea (rrP-) C. walnut (R-P-) D. single (rrpp) R-P- = walnut (9) R-pp = rose (3) 62 rrP- = pea (3) rrpp = single (1) Chicken Comb-type Gene Interaction example: • 1) Phenotypic ratio in F2 remains unaltered from classical ratio. • 2) One distinct phenotype in each phenotypic class-rather than two, although two separate genes are being considered. • 3) The appearance of “novel” or new phenotypes in the F1 and/or F2 generations that had not appeared in the P generation. Illustrate gene interaction for comb-type in chickens by filling in all blanks below. Circle all gametes. R-P- = walnut; R-pp = rose; rrP- = pea; rrpp = single P Phenotypes walnut Genotypes RRPP Gametes 1/1 RP x x x single rrpp 1/1 rp . F1 Phenotypes walnut x walnut Genotypes RrPp x RrPp Gametes 1/4 RP 1/4 Rp 1/4 rP 1/4 rp x 1/4 RP 1/4 Rp 1/4 rP 1/4 rp F2 Genotypes F2 Phenotypes F2 Phenotypic ratio F2 Genotypic ratio Working space Working space 9:3:3:1 4:2:2:2:2:1:1:1:1 . . .: .: . . . . USE THE INFORMATION ABOVE TO MAKE A TESTCROSS BELOW F1 Parent Testcross Parent Phenotypes walnut Genotypes RrPp Gametes 1/4 RP 1/4 Rp 1/4 rP 1/4 rp x x x single rrpp 1/1 rp Testcross Progeny Phenotypes 1/4 walnut 1/4 rose . Genotypes 1/4 pea 1/4 single Testcross Phenotypic Ratio Testcross Genotypic Ratio 1/4 RrPp 1/4 Rrpp 1:1:1:1 1:1:1:1 63 1/4 rrPp 1/4 rrpp . . . . . Illustrate gene interaction for comb-type in chickens by filling in all blanks below. Circle all gametes. R-P- = walnut; R-pp = rose; rrP- = pea; rrpp = single P Phenotypes rose Genotypes RRpp Gametes 1/1 Rp x x x pea rrPP 1/1 rP . F1 Phenotypes walnut x walnut Genotypes RrPp x RrPp Gametes 1/4 RP 1/4 Rp 1/4 rP 1/4 rp x 1/4 RP 1/4 Rp 1/4 rP 1/4 rp F2 Genotypes F2 Phenotypes F2 Phenotypic ratio F2 Genotypic ratio Working space Working space 9:3:3:1 4:2:2:2:2:1:1:1:1 . . .: .: . . . . USE THE INFORMATION ABOVE TO MAKE A TESTCROSS BELOW F1 Parent Testcross Parent Phenotypes walnut Genotypes RrPp Gametes 1/4 RP 1/4 Rp 1/4 rP 1/4 rp x x x single rrpp 1/1 rp Testcross Progeny Phenotypes 1/4 walnut 1/4 rose . Genotypes 1/4 pea 1/4 single Testcross Phenotypic Ratio Testcross Genotypic Ratio 1/4 RrPp 1/4 Rrpp 1:1:1:1 1:1:1:1 64 1/4 rrPp 1/4 rrpp . . . . . Novel phenotypes in the F2 as well as modified F2 dihybrid phenotypic ratios can be seen in summer squash (Cucurbita pepo) fruit shape. This is in contrast to chicken comb-type, which only produces novel phenotypes and not modified F2 phenotypic ratios. Ex. Classical F2 ratio 9 3 3 1 Genotype Phenotype A_B_ A_bb aaB_ aabb disc sphere sphere long 65 Final F2 Phenotypic ratio 9 disc 6 sphere 1 long
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