GENETICS WORKSHEET ANSWER SHEET

Transcription

1 GENETICS WORKSHEET ANSWER SHEET NOTE: Pages 1 & 2 are not included here since there are no problems on these pages. Section I: Monohybrid Crosses 1. In tomatoes, red fruits are dominant over yellow fruits. A. What letter would you choose to represent the red and yellow alleles. RED ALLELE = R YELLOW ALLELE = r 2. Using your answers from #1, determine the genotypes of these organisms (remember, 2 letters for each trait). A. homozygous red fruited plant = RR. heterozygous plant = Rr C. yellow fruited plant = rr D. What is the phenotype (appearance) of plant? RED E. Why is it unnecessary to call plant C homozygous yellow fruited? yellow fruit is the recessive phenotype - must be homozygous recessive 3. For each of the answers in #2, list all the unique gametes that could be produced. A. R. R & r C. r USE THE INFORMATION IN #1 - #3 AOVE TO SOLVE THIS PROLEM: 1. If you cross a homozygous red-fruited plant with a yellow-fruited plant, what is the appearance (phenotypes) and genotypes of the F generation? 1 RR x rr all progeny are Rr - red-fruited 2. What will be the phenotypes and genotypes of the F 2 generation. Rr x Rr Phenotype - 3 Red : 1 yellow Genotype - 1 Homozygous dominant : 2 Heterozygous : 1 Homozygous recessive. 25Apr13 2:14 pm -1-

2 MORE PROLEMS: 3. W = White w = yellow Male = Ww Female = ww W w w Ww ww Phenotype - 1 White : 1 yellow Genotype - 1 Heterozygous : 1 Homozygous recessive 4.! Determine the genotypes of all seven animals. H = Hornless h = horned ull: Hh Cow A: hh Cow : Hh Cow C: hh Calf D: hh Calf E: hh Calf F: Hh 5. When heterozygotes mate the resulting phenotypic ratio is always 3 dominant phenotypes to 1 recessive phenotype. Therefore, you would expect 3:1 ratio of brown-haired children to blonde-haired children 6. F = farsighted f = normal vision F f normal man = ff f Ff ff farsighted woman with normal father (ff) = Ff S Woman must be heterozygous since her father can only contribute a "f" to his progeny. S ½ of the offspring would have normal vision 25Apr13 2:14 pm -2-

3 7. Short haired male x short-haired female #1 = all short-haired offspring x short-haired female #2 = 3:1 short:long haired offspring Short hair (S) dominant to long hair (s) 1st cross: 2nd cross: all progeny have at least one dominant allele (S ). Therefore both parents are either homozygous dominant or one is a heterozygote while the other is homozygous dominant. A 3:1 phenotypic ratio indicates heterozygous parents. Therefore both parent (the male and female #2) must be heterozygous. ased on the results of cross #2 the first female must be homozygous dominant since the male is heterozygous. 8. T = trotter t = pacer Trotter (T ) x pacer (tt) Progeny are in approximate 1:1 ratio. Since pacer progeny were produced the Trotter parent must carry a recessive allele. Original parents: Trotter - Tt pacer = tt 9. Normal feathers = N silky feathers = n Two heterozygotes would yield 3 normals to 1 silky. If 96 birds were raised then 3/4 would have normal feathers and 1/4 would have silky feathers. 3/4 of 96 = 72 1/4 of 96 = 24 Section 2: DiHybrid Crosses In peas, a single gene codes for seed shape and another single gene codes for stem length. Each gene has two alleles, one dominant and one recessive. For stem length, tall plants are dominant over short plants. For seed shape, smooth peas are dominant over wrinkled peas. A. Choose letters to represent each gene and its alleles. Seed shape: Smooth: S Stem length: Tall: T wrinkled: s short: t 25Apr13 2:14 pm -3-

4 Table of Genotypes for Two Loci Phenotype Possible Genotypes Possible Unique Gamete Genotypes SMOOTH, TALL SSTT ST SsTT ST st SSTt ST St SsTt ST st St st SMOOTH, short SStt St Sstt St wrinkled, TALL sstt st st sstt st wrinkled, short sstt st st 1. SSTT x sstt SSTT gametes = ST sstt gametes = st All offspring will be SsTt 25Apr13 2:14 pm -4-

5 2. What are the F 2 genotypes and phenotypes (crossing two F 1 from problem #1) ST st St st ST SSTT SsTT SSTt SsTt st SsTT sstt SsTt sstt St SSTt SsTt SStt Sstt st SsTt sstt Sstt sstt 9 Smooth Tall : 3 wrinkled Tall : 3 Smooth short :1 wrinkled short MORE PROLEMS: 3. H = hairless h= hairy L = long wings l = vestigial wings male = hhll female = HHLL male gametes = hl female gametes = HL ALL OFFSPRING = HhLl HL hl Hl hl HL hl Hl hl HHLL HhLL HHLl HhLl HhLL hhll Hairy Long HhLl hhll Hairy Long HHLl HhLl HHll Hairless vestigial Hhll Hairless vestigial HhLl hhll Hairy Long Hhll Hairless vestigial hhll hairy vestigial 25Apr13 2:14 pm -5-

6 4. = ucktoothed b = snaggletoothed M = Matted hair m = frizzled hair bbmm x MM 9 - ucktoothed Matted hair 3 - snaggletoothed Matted hair 3 - ucktoothed frizzled hair 1 - snaggletoothed frizzled hair 5. = lack Fur b = brown fur N = Normal length n = short fur ALL OFFSPRING ARE bmm lack, Normal = N X brown, short = bbnn THIS IS A TEST CROSS PROLEM N x bbnn = 1:1 ratio black,shorts to black,normals Since all offspring are black, the black parent must be homozygous dominant Since half the offspring are short and the other half are normal the normal parent must be heterozygous 6. = excretes betamin b = does not excrete betamin M = does not excrete methanethiol m = excretes methanethiol oth parents = M Son = bbmm Parents must be double heterozygous (bmm) since this is the only pairing that can yield a double homozygous recessive offspring. There was a 1/16 chance of this child being born. 9 - excrete etamin, DO NOT excrete methanethiol,= M 3 - DO NOT excrete etamin, DO NOT excrete methanethiol= bbm 3 - excrete etamin, excrete methanethiol = mm 1 - DO NOT excrete etamin, excrete methanethiol = bbmm 7. Singer, smelly x non-singer, fragrant yields all NON-SINGERS, SMELLY This data tells you that NON-SINGER and SMELLY are the dominant traits S = smelly N = non singer s = fragrant n = singer singer, Smelly = nnss Non-singer, fragrant = NNss 25Apr13 2:14 pm -6-

7 Section 3: Incomplete Dominance 1. R=Red W=White RR x WW All offspring are RW = pink 2. RW x RW 3. R = Red W = White RW = Roan H = Hornless h = horned R W R RR RW W RW WW 1:2:1 phenotype and genotype ratio 1 Red:2 Pink:1White A. hhww x HHRR All offspring will be HhRW = hornless and roan. HR HW hr hw HR HHRR HHRW HhRR HhRW HW HHRW HHWW HhRW HhWW hr HhRR HhRW hhrr hhrw hw HhRW HhWW hhrw hhww For Mendelian trait = 3 hornless : 1 horned (12:4) For Incomplete trait = 1 red : 2 roan : 1 white (4:8:4) 25Apr13 2:14 pm -7-

8 Section 4: Sex-Linked Genes b b 1. Woman = X X Man = X Y b X X X X b b Y X Y ALL MALE OFFSPRING ARE COLORLIND ALL FEMALE OFFSPRING ARE NORMAL UT CARRY THE COLOR-LIND GENE H h 2. Normal woman = X X (Father was hemophiliac so normal daughter would be heterozygous) Normal man = X H Y Section 5: Multiple Alleles X H X H H H H h X X X X X h H h Y X Y X Y Female offspring: 50% homozygotes; 50% heterozygotes All Normal Male offspring: 50% hemizygous H ; 50% hemizygous h 50% normal; 50% hemophiliac All offspring: 75% Normal; 25% hemophiliac A 1. A Father = I I O Mother = ii I I A i A I i I i 25Apr13 2:14 pm -8-

9 A 2. Heterozygous A father = I i Heterozygous mother = I i I A A A I I I I i i I i ii OFFSPRING = 1 A : 1 A : 1 : 1 O i A 3. A parent = I I O parent = ii I A A i I i I i Child with lood type A: 50% or 1 in 2 4. 'Your' father's and 'your' blood type: ii I A Possible genotypes of mother: I i, I i, ii - Mother must have at least one i allele to produce a Type O child Could these same parents have a child with blood type A? Explain - No, since the father would have to carry the A or allele to produce an A child. The mother would have to be heterozygous as well. A A A 5. Type A person, possible genotypes: I I or I i Type person, possible genotypes: I I or I i A YES, provided both parents are heterozygotes (I i and Ibi) 25Apr13 2:14 pm -9-

10 Section 6: Calculating % Recombination of Linked Genes. 1. EeRr x eerr (ELONGATED/RED x rounded/white) - UNLINKED ER er Er er 50% Parental (P) 50% Recombinant (R) LOCI ARE LINKED re EeRr ELONGATED/RED (P) eerr rounded/red (R) Eerr ELONGATED/white (R) eerr rounded/white (P) Number of Offspring out of 1000 total ELONGATED/RED (P) 370 rounded/white (P) 125 rounded/red (R) 125 ELONGATED/white(R) What percentage of the offspring are parental? recombinant? 1000 total offspring 750 Parental = 75% 250 Recombinant = 25% - 25% Recombination frequency 2. a. GgSs X ggss UNLINKED of each phenotype type b. 740 GREEN-LASTING, HAVE STINGER (P) 640 red-blasting, stingerless (P) 300 GREEN-LASTING, stingerless (R) 320 red-blasting, HAVE STINGER (R) Total Offspring (P)arental = 1380/2000 = 69% (R)ecombinant = 620/2000 = 31% = Recombination Frequency 25Apr13 2:14 pm -10-

11 Section 7: Mapping Genes using Recombination Frequencies A-Z = 12% Q-F = 37% Q-Z = 22% A-Q = 10% NOTE- MAP NOT TO SCALE Step 1 = Map loci with greatest Recombination frequency (greatest map distance) Step 2 = Map Z in relation to Q Step 3 = Map A in relation to Z & Q Additional helpful recombination frequencies for validation 1. Z-F (validate predicted distance of 15) 2. A-F (validate predicted distance of 27) 25Apr13 2:14 pm -11-