GENETICS WORKSHEET ANSWER SHEET

1 27 May10 4:01 pm-1- GENETICS WORKSHEET ANSWER SHEETNOTE:Pages 1 & 2 are not included here since there are no problems on I: Monohybrid tomatoes, red fruits are dominant over yellow What letter would you choose to represent the red and yellow ALLELE = R YELLOW ALLELE = r your answers from #1, determine the genotypes of these organisms (remember, 2letters for each trait).A. homozygous red fruited plant = RR B. heterozygous plant = Rr C. yellow fruited plant = rr D. What is the phenotype (appearance) of plant B ? RED E. Why is it unnecessary to call plant C homozygous yellow fruited ? yellow fruit is the recessive phenotype - must be homozygous each of the answers in #2, list all the unique gametes that could be R B. R & r C. r USE THE INFORMATION IN #1 - #3 ABOVE TO SOLVE THIS you cross a homozygous red-fruited plant with a yellow-fruited plant, what is the1appearance (phenotypes) and genotypes of the F generation ?

2 RR x rr all progeny are Rr - red-fruited22. What will be the phenotypes and genotypes of the F x RrPhenotype - 3 Red : 1 yellowGenotype - 1 Homozygous dominant : 2 Heterozygous : 1 Homozygous 4:01 pm-2-MORE squash, white color is dominant over yellow. Pollen (male gametes) from aheterozygous white fruited plant is placed on the pistil (female organ) of a yellow fruitedplant. Determine the genotypes of the parents and the genotypes and phenotypes of1the F = Whitew = yellowMale = Ww Female = wwWwwWwwwPhenotype - 1 White : 1 yellowGenotype - 1 Heterozygous : 1 Homozygous hornless bull is bred to three cows, A,B, and C. Cow A is horned and produces Calf D, which isalso horned. Cow B is hornless and produces Calf E, which is horned. Cow C is horned andproduces Calf F which is hornless. Note: Hornless is dominant to horned.! Determine the genotypes of all seven animals. H = Hornlessh = hornedBull: Hh Cow A: hh Calf D: hh Cow B: Hh Calf E: hh Cow C: hh Calf F: Hh gene for brown hair is dominant over that for blond hair.

3 If two heterozygotes married, whatproportions of these two traits would you expect in their children ?When heterozygotes mate the resulting phenotypic ratio is always 3 dominant phenotypes to 1recessive phenotype. Therefore, you would expect 3:1 ratio of brown-haired children to blonde-haired children27 May10 4:01 humans the gene for farsightedness is inherited as a dominant (therefore normal vision isrecessive). What fraction of children will have normal vision if a normal man marries a woman whois farsighted and had a normal father ?F = farsightedf = normal visionnormal man = fffarsighted woman with normal father (ff) = Ff SShe must be heterozygous since her father can only contribute a "f" to his of the offspring would have normal short-haired female guinea pigs are mated to a short-haired male guinea pig. Several littersare produced. The first female produces 25 short-haired offspring and no long-haired ones. Thesecond female produces 15 short-hairs and 5 long-hairs.

4 Based on this information, whatdeductions can be made concerning hair-length inheritance ? List the genotypes of all the parentsand haired male x short-haired female #1 = all short-haired offspring x short-haired female #2 = 3:1 short:long haired offspringShort hair (S) dominant to long hair (s)1st cross:all progeny have at least one dominant allele (S__ ). Therefore both parents areeither homozygous dominant or one is a heterozygote while the other ishomozygous cross:A 3:1 phenotypic ratio indicates heterozygous parents. Therefore both parent(the male and female #2) must be on the results of cross #2 the first female must be homozygous dominant since the maleis heterozygous. horses, trotter is dominant over pacer. A trotter is mated to a pacer. Over the course of severalmatings, 5 pacers are produced along with 4 trotters. What were the genotypes of each originalparents ?

5 T = trotter t = pacerTrotter (T___ ) x pacer (tt) Progeny are in approximate 1:1 ratio. Since pacer progeny were produced the Trotter parentmust carry a recessive parents: Trotter - Tt pacer = tt27 May10 4:01 feathers in fowl is recessive to normal feathers. If 96 birds were raised from a cross betweentwo heterozygotes, how many would you expect to be normal ? silky ?Normal feathers = N silky feathers = nTwo heterozygotes would yield 3 normals to 1 silky. If 96 birds were raised then 3/4 would havenormal feathers and 1/4 would have silky of 96 = 721/4 of 96 = 24 Section 2: DiHybrid CrossesDihybrid crosses examine two unlinked (on different chromosomes) gene loci. The following problemuses many of the skills you learned in the section on monohybrid peas, a single gene codes for seed shape and another single gene codes for stem length . Eachgene has two alleles, one dominant and one recessive. For stem length, tall plants are dominant overshort plants.

6 For seed shape, smooth peas are dominant over wrinkled letters to represent each gene and its shape: Smooth: S wrinkled: s Stem length: Tall: T short: t the Table on the next page is a list of all the possible phenotypes for this pair of traits. Determine the genotypes for these plants, in some cases there will be multiple possiblegenotypes. When we worked with one gene locus, each individual genotype was representedby 2 letters since individual organisms are diploid. Now we are working with two gene loci, soeach individual genotype will be represented by 4 letters, 2 letters for each gene locus. Fill inthe first part of the Table on the next page with all the possible genotypes for each phenotypelisted. The first genotype is each of the individuals in part B, list all the unique gametes that can be produced. Remember, each gamete will have 1 allele from each gene locus. When we worked with asingle gene, each gamete genotype was represented by 1 letter.

7 Now we are working with 2gene loci, so each gamete genotype will be represented by 2 letters, one from each gene in the rest of the Table on the next page with all the possible gamete genotypes for eachparent genotype listed. The first gamete genotype is 4:01 pm-5-Table of Genotypes for Two LociPhenotypePossibleGenotypesPossible UniqueGamete GenotypesSMOOTH, TALLSSTTSTSsTTSTsTSSTtSTStSsTtSTsTStstSM OOTH, shortSSttStSsttStstwrinkled, TALLssTTsTssTtsTstwrinkled, shortssttstUSE THE TABLE OF INFORMATION TO SOLVE THE NEXT 2 the F genotypes and phenotypes resulting from a cross of a homozygous Smooth andhomozygous Tall plant with a wrinkled, short x ssttSSTT gametes = STsstt gametes = stAll offspring will be SsTt27 May10 4:01 are the F genotypes and phenotypes (crossing two F from problem #1)STsTStstSTSSTTSsTTSSTtSsTtsTSsTTssTTS sTtssTtStSSTtSsTtSSttSsttstSsTtssTtSstts stt9 Smooth Tall : 3 wrinkled Tall : 3 Smooth short :1 wrinkled shortMORE fruit flies, the genes for wing length and body hair each have two alleles.

8 Long wings and ahairless body are dominant to alleles for vestigial (functionally useless) wings and a hairy the genotypes and phenotypes of the offspring from the cross between a hairy,vestigial winged male and a homozygous dominant = hairless h= hairy L = long wings l = vestigial wingsmale = hhllfemale = HHLL male gametes = hlfemale gametes = HLALL OFFSPRING = the genotypes and phenotypes of the F LongHhLLHairless LongHHLlHairless LongHhLlHairless LonghLHhLLHairless LonghhLLHairy LongHhLlHairless LonghhLlHairy LongHlHHLlHairless LongHhLlHairless LongHHllHairlessvestigialHhllHairless vestigialhlHhLlHairless LonghhLlHairy LongHhllHairlessvestigialhhllhairy vestigial27 May10 4:01 snerds, matted-hair is dominant to frizzled-hair and buck-toothed is dominant to snaggle-toothed. frizzled-hair, snaggle-tooth snerd is mated to a pure breeding matted-hair, buck-toothed1snerd. What is/are the genotypes and phenotypes of the = Bucktoothedb = snaggletoothedM = Matted hairm = frizzled hairbbmm x BBMMALL OFFSPRING ARE doing any calculations, what will be the phenotype ratio of the - Bucktoothed Matted hair 3 - snaggletoothed Matted hair3 - Bucktoothed frizzled hair1 - snaggletoothed frizzled rabbits, black fur and normal length fur are dominant to brown fur and short fur.

9 The litter from amating of a black, normal fur rabbit with a brown, short rabbit contains 6 black, shorts and 7 black,normals. What are the genotypes of the parents ? B = Black Furb = brown furN = Normal lengthn = short fur Black, Normal = B__N__brown, short = bbnnTHIS IS A TEST CROSS PROBLEMB__N__ x bbnn = 1:1 ratio black,shorts to black,normalsSince all offspring are black, the black parent must be homozygous dominantSince half the offspring are short and the other half are normal the normal parent must beheterozygous27 May10 4:01 eating asparagus, some individuals excrete the strongly odorous substance, methanethiol. The excretion of this substance is due to a recessive gene. After eating red beets, someindividuals excrete the red pigment, betamin. This is due to a dominant gene. A man and awoman both excrete betamin but do not excrete methanethiol. Their son excretes methanethiol butdoes not excrete betamin. Give the genotypes of the man, woman, and their son.

10 What genotypeand phenotypes and in what proportions can be expected in subsequent progeny ?B = excretes betaminb = does not excrete betaminM = does not excrete methanethiolm = excretes methanethiolBoth parents = B__M__Son = bbmmParents must be double heterozygous (BbMm) since this is the only pairing that can yield adouble homozygous recessive offspring. There was a 1/16 chance of this child being born. 9 - excrete Betamin, DO NOT excrete methanethiol,= B__M__3 - DO NOT excrete Betamin, DO NOT excrete methanethiol= bbM__3 - excrete Betamin, excrete methanethiol = B___mm1 - DO NOT excrete Betamin, excrete methanethiol = the planet Zion, you have discovered two interesting plants. These plants exhibit several traits,two of which you decide to study genetically. The two traits are singing (production of a perfect "C"note upon touching) vs. non-singing and smelly (exude a odor of manure) vs. fragrant (exude anodor similar to earth roses).