‘Name Period Codominance Worksheet (Blood types) Human blood types are determined by genes that follow the CODOMINANCE pattern of inheritance ‘There are tivo dominant alleles (A & B) and one recessive allele (0). - Biood Type] Genotype Tan donate blood Can reesive blood rom: |_ (Phenotype) E ° (00) ABABandO f+ ° (universal donory_-— AB ne 8 ‘AB,AB and O z (universal receiver) A ebor i | ABA ~ OA roy — B PP or 185 ABB op 0) ‘Write the genotype for each person based on the description: Homozygous forthe “B” allele . Heterozygous for the “A allele oe c. TypeO 2 4. Type “A” and had a type “O” parent e. Type“AB” = £. Blood can be donated to anybody —_ Can only get blood from a type “O" donor oO © 2, Pretend that Brad Pit is homozygous for the type B allele, and Angelina Jolie is type “O. ‘What are all the possible blood types of their baby? (Do the purnett square) © taD® are SRowing all the possible blood types forthe offspring produced | py 3. Complete the punnett sq by type “0 mother and ana Type "AB" father. What are percentages of each offspring? 60Lhype A 50% haps B_ 8 50." They have three children named Matthew, 4. Mrs, Essy is type “A” and Mr. Essy is type “A,” and Luke is ype “AB.” Based on Mark, and Luke. Mark is type “O," Matthew i type this information: 2 Mr Exsy mast ave he snaps QO 'b. Mrs. Essy must have the genotype | because [ACCIK has blood type, One cannot be the child ofthese parents because nee parent has the allele ‘Two parents think their baby was switched atthe hospital. ts 1968, so DNA fingerprinting technology “Joes not exist yet. The mother has blood type “O,” the father has blood type “AB,” and the baby has blood type “B.” Father's genotype: Be Baby's genorype aan aaa showing ail possible genotypes for children produced by this couple. Was the baby switched? ‘Mother's nope: ©: A ° BoBLOOD TYPE & INHERITANCE i blood typing, the gene for type A and the gene for type Bare codominant. The gene for ype O is recessive, Using Punnett squares, determine the possible blood types of the offspring when: 1. Father is type 0, Mother is type O ©| 66 | co Go-*e o| 00 | oe = 2. Father is type A, homozygous; Mother is type B, homozygous A * Bl AB | re oe in| sie % B Bl AB Ke CO _% AB 3, Father is type A, heterozygous; Mother is type B, heterozygous A 0 ZS %o b|AB | Bo tons 26% 88 0 [Av | 00 | 4, Father is type O, Mother is type AB mee Oo _% AB 5, Father and yea are bef tvne AB —o__%0 fw 7 AB 25% A -— eat B| AP | ee ee1. Practice setting up keys for the phenotypes listed in each set. Remember that the "medium" trait must always be heterozygous. a) Birds can be blue, white, or white with blue-tipped feathers. BB = blue; BW = blue tips; WW = white OR you can use other way to set up problems using a single letter BB = blue, Bb = tips, bb = white b) Flowers can be white, pink, or red. RR = red, RW = pink, WW=white c) A Hoo can have curly hair, spiked hair, or a mix of both curly and spiked. CC = curly, CS = mixed, SS = spiked d) A Sneech can be tall, medium, or short. TT= tall, TS = medium, SS = short e) A Bleexo can be spotted, black, or white. BB = black, BW = spotted, WW = white 2. Now, can you figure out in the above list, which of the letters represent codominant traits and which are incomplete. if it "BLENDS* then it is incomplete dominance, if both traits are expressed, it is codominant. Students can become overly anxious about this difference, when in fact the problem solving is no different and the difference is largely a matter of semantics. __ incompletely Dominant bd Codominant __a,, € 3. In Smileys, eye shape can be starred, circular, or a circle with a star. Write the genotypes for the pictured phenotypes OO ‘<7 §§ = star, SC = circle-star, CC = circle 4, Show the cross between a star-eyed and a circle eyed. SS x CC What are the phenotypes of the offspring? __all star circle. What are the genotypes?__ SC.Simple Genetics Practi roblems (10 points) 1. For each genotype, indicate whether it is heterozygous (HE) or homozygous (HO) AA be He Wi Mm te. nT wos xk ic ba He, BH HG. Fis meee 2. For each of the genotypes below, determine the phenotype. Purple flowers are dominant 10 white flowers | Brown eyes are dominant to blue eyes PP Ne — BB — Pp Bb —_ PP te = bb Woke so ‘Rowad seeds arg dominant to wrinkled Boal ars recessive (long tails dominant) RR. Rr Bass UN Saeees 3. For each phenotype, list the genotypes. (Remember to use the letter of the dominant trait) Sraigh jars dominanttoeury. | Poligd heads are dominant to rowed heads TA aight pee ind fii saan ~ Poi =e Found 4. Set up the square for each of the crosses listed below. The trait being studied is round seeds (dominant) and wrinkled seeds (recessive) erat < « [kv vv VY | Re jive What percentage of the offspring will be round? St OG21ee | er v ~ i a 6. A Ttplant is crossed with a Tt plant <0 faa M © Aya percentage ofthe affapring wil be shor? aL etka 20°L RrxRr v er | vv ‘What percentage of the offspring will be round? KL mye ee | ee er | be What percentage of the offspring will be round? ACOZ.. Practice with Crosses, Show all work! 5 ATT (tall) plant is crossed with at (short pland).y yncy ‘What farecnage ofthe offspring wil etal? - WOLSLS a 7. A heterozygous round seeded plant (Rr) is erossed with a homozygous round seeded plant (RR). What percentage of the offspring will be homozygous (RR)? _ 8. A homozygous round seeded plant is crossed with a homozygous ‘wrinkled seeded plant. What are the genotypes of the parents? AR x 0c ‘What percentage of the offspring will also be homozygous? Fy 7. Vv \erl ee 91m p plants pup lower ate dominant to white lowers Veen setae sais Towered plants ae eross, what pergentage of their cert See oe toa wo ¥ ~ Ire Ne ala icevoy 10. A white flowered plant is erossed wit th a plant that is heterozygous for the trait. What percentage ofthe e Es offspring will have purple flowers? ota Y Iyydyr 11. Two plants, both heterozygous f : gous for the gene that controls flower color are crossed. What percentage of their offspring g ee will have purple flowers? =" / : Valet What petcentage will have white flowers? 2S RIA) 12. In guinea pigs, the allele for short hair is dominant. What genotype would a heterozygous short haired guinea pig have? A Oe What genotype would a purebreeding short haired guinea pig have? ‘What genotype would along haired guinea pig have?_Q O— 13. Show the cross for a pure breeding short haired guinea pig A A and a long haired guinea pig. oY What percentage of the offspring will have short hair? 1007, a \Aa 14, Show the cross for two heterozygous guinea pigs. a tAa What percentage of the offspring will have short hair? Q ‘What percentage of the offspring will have long hair? 15. Two short haired guinea pigs are mated several times. Out of 100 offspring, 25 of them have long hair. What are the probable genotypes of the parents?_A fx x ___ AX /A__ Show the cross to prove it! AoGenetics - x Linked Genes In fruit fi NS: ee color is a sex linked trait. Red is dominant to white.** 1. What are t © the sexes and eye colors of flies with the following genotypes? XE becd x8y Us fod eee Eola cde xR To ry Me xB Red ery M y Whike 2. What are the genotypes of these flies: white eyed, male XY red eyed female (heterozygous) xe white eyed, female X" XY" red eyed, male Xay eyed female X 'X ‘with a red-eyed male X® Y. «xe ve 4, Show a cross between a pure red eyed female anda white eyed male ‘What are the genotypes of the parents: Xf Xx R r é XEXF oa XY xh How many are: y ex? white eyed, male_© /. white eyed, female C)'f red eyed, male_7 red eyed, female [50 7. 5. Show the cross of a red eyed female (heterozygous) and a red eyed male. ‘What are the genotypes of the parents? a kyr R x 4“ XX 12 XP s xX @ x How many are: white male S. moamee, = oY key ley | red eyed, male red eyed, female =)",Human Sex Linkage 6: In humans, hemophilia is a sex linked trait. Females can be normal, carriers, or have the disease. Males will either have the disease or not (but they won’t ever be carriers) X"'X "= female, normal X HY = male, normal x x X"X = female, cartier L female, hemophiliac X= male, hemophiliac x Nats by! X"X Show the cross of. man who has hemophilia with a woman who is a carrier. Y HY Nas ‘What is the probability that their children will have the disease? 7..A woman who is a carrier marries a normal man. Show the cross. What is the probability that their children will have hemophilia? What sex will a child in the family with hemophilia be? 8, A woman who has hemophilia marries a normal man. How many of their children will have hemophilia, and what is their sex? a yh 4. ny) OST. Chance peseyex cB oe xe yh i g. bps SU Chaas® \ abe aeName. Period, C Worksheet: Dihybrid Crosses UNIT3:GENETICS Determine what kind of problem you are trying to solve. Determine letters you will use to specify traits. Determine parent's genotypes. Make your punnett square and make gametes Complete cross and determine possible offspring. Determine genotypic and phenotypic ratios. Two-Factor Crosses (Di-hybrid) Ex) Atall green pea plant (TTGG) is crossed with a short white pea plant (tteg). TrorTt=tall tt=short GGorGg=green ——gg= white 16 16 Te 16 TiGe [Te | TGs [te Tig | TtGg | tee | TtGe tg te [Te | Tee Ties | TtGe te te [Tteg | Tee TtGg | TtGg hike: O'Short/Green OShort/ White 6 Tall/Green: OTal 1) Atal green pea plant (77Ge) is crossed witha tll green pea plant (Ge) WGig_ x WAC 3 Cc Ng ey VG [TES | Wg [TIO TT 5 9 | Waa |T1bg |"135 IT1Gg [Tae AS Iraiycreen QS raywnite: OL snor/Green: C/short/ white IL lb ° 0Period, Name. —__ it a plant (tteg). 2) Atall green pea plant (TtGg) is crossed with a Short white pea pl Thag 1 ktgg VG & 45 ely Re, ees ah kg [ikag [Te [tg jtha4 Ms [Rs Ming feecsrns | 4 fay Hens ec es 25fraycreen QSL taivwhite :95 J short/Green : 25/short/ white 4 4 4 4 3) AHomozygous tall, green flowered plant is crossed with a Homozygous short white flowered plant. TIGG x ityg VG TG re 4g [feta [ees [eG [rec] 44 feGy MTG [Teag [Teas | 63 pees Gg pees r by Meg Heaq Gy [QL raivercen : CY.ratpwnite - OL, shore/gcen ALsnorywnie + IG 0 © oOName —_———— Period 4) Two Heterozygous Tall, Green pea plants are crossed. Thbg x Tkhg Ta *tG 49 TG [T1aG | Treg TREC |e T4 [TTGg [TTAS [THOS [eas AG [Teac [reds head bet Ao, [Teas [Tea heeea [tea] taiyereen: 2 frauywnite Sfroncieen A forts write Vo fu lb I 4. In man, assume that spotted skin (S) is dominant over non-spotted skin (s) and that wooly hair (W) is dominant over non-wooly hair (w). Cross a marriage between a heterozygous spotted, non-wooly man with a heterozygous wooly-haired, non-spotted woman. Give genotypic and phenotypic ratios of offspring. S oe Wi Sw wd Sw Se Ssum ¥ S58 a Ssh | Seka] 5 [ssa s' 251s se aise Weoly har sW |Ssho sso ssh] 25 1.{4/1») non Speed sv [Ss si0u 28 7, (A) te) seatied Nor—atly 5 7. non -Seott. dj Woll. 2. Inhorses, black is dependent upon a dominant gene, B, and chestnut upon its recessive 2 tS a eect dominant gore, Tine pacing gat tote recessive ale, ene et oats mated a homosygeus chestnut weer, what wi be the arte ove a? gett: Wir _ et gt et st Bore | ROTH] RTE] Ate ; bot | Bore [abre [Bye | Bott \OO/, Black — be [Burr [Te [ete [ere otters ob | Bat | Rote [6bTE TEpies —\ \ disk-shaped frurt it color (w) and ‘ (Ww) tie cara for white, disk-shaped fruit 3. in summer squash, white ie na a vay, Wa ssh plant ee ‘ rant over sphere-snaped ful uit, . fearon etn SIN, WWD ee SL \00!. see are we Bow ake meas ist abd ed et wire Fe wd {WDA WDA Nb WH Soest Ho DA aod adda c He 5 nt The recessive tat causes mice to run in circles OF vice with this trait are called ir Tieng mice (t), Haircolor is also inherited in mice, Black Hav (B) is dominant over brown hs a) For each of the following problems, determine the parent ‘genotypes, determine poss! igemetes then construct a Punnet square to solve, «a, Cross @ heterozygous running, heterozygous biack mouse with a homozygous running, homozygous black mou: Parente geratypes Fur Wo X RABE Peaabi games eo eo ES re x AHRG RS RS ’ Re AB RB Re \Og/. $& (Pee|eres| Pees | ere! Running Boot Rb [ARB ReBd | CRED RRAd ® lee po ee eb PE Vb (Rrbn| PrBbo| RrBb RB, b crea oar Cane back mouse wih. elon gous runing, brown Tea cenotyoes BRAD x Rr oat ve” Rance SNARE ee" be 08 evel PER - ee [PLE | AEB) Kepel PEE Il. Rurninas! ev [Pee ref peep Bade 1 ee ErOb |e cB] Oy ep vp [PrB> [Pré | er Bp| ey Bb 6 Cross a waltzing brown mouse with a waltzing brown mousePeriod _— Parental gent b earttcensyoes yr Lb tee nas lesb [rrbiy | rib || 2 ee lrreb 1O0/ emt ae vyh | feo op rie | ne ebb | Trev vb rob rb | yreb vroh! 09 ‘Offspring phenotypic ratio 4d, Cross a homozygo4 ir. heteroryaous black mouse with a waltzing brown mouse Parental genotypes. Possible gametes be Offspring phenotypic ratio ~All ie [BrGin | Rebbo! RB) Erb : Ruri BO). Rareningf OO Vy [Bab [bee 04 Rete Bol. [, Ronnie! Po | ReBr [Arb | Pro F Reo re [eats Brie | P<6y rbd sper smn tema ina ni —— soc © sgensyees Reb x Ke BE Ro yo ve Perea al genotype x Re as [Reto] PBL RR) eo SNe seo oat ‘eel reed 8 [PeBio Rr Be \erB> | HB Bla _v6 [ete ts se rg heterozygous black 8 Blo a 26 _ We eS _ b Parental genotypes fo Possibie gametes ____ ‘Offspring phenotypic ratio “S Qucning Bae 2 Prurrg) 2 Ra wag Bes rT Rhack = SanaSILana