Genetics Worksheet Answers

Genetics Worksheet Name___________________ Section A: Vocabulary 1. Identify if the alleles are homozygous (Ho) or heterozygous (He). a. DD Ho b. Ee He c. tt Ho d. Hh He 2. For each genotype below, determine the phenotype. a. Purple flowers are dominant to white flowers. PP purple b. Pp purple c. pp white 3. For each phenotype, identify the genotype. a. Curly hair is dominant to straight hair. What are the possible alleles if you have curly hair? Cc, CC What are the possible alleles if you have straight hair? cc 4. Fill in the chart with the missing information. R = right handed r = left handed Genotype Phenotype Homozygous RR Right handed X 5. Rr Right handed rr Left handed Heterozygous x Fill in the missing vocab based on the definition below. a. RECESSIVE allele from that is only seen when it is homozygous b. ALLELES alternative versions of genes c. HOMOZYGOUS/PURE these alleles are the same d. HETEROZYGOUS/HYBRID these alleles are different e. PURE another name for homozygous f. PHENOTYPE the physical appearance of a trait g. GENOTYPE an organisms alleles Section B: Monohybrid Crosses 1. Round seeds (R) are dominant to wrinkled (r) seeds. Cross a homozygous dominant seed with a heterozygote. Parents = RR x Rr R R Genotype: 50 % RR 50 % Rr 0 % rr RR RR R Phenotype: 100 % round 0% wrinkled Rr Rr r 2. A black coat is dominant over brown coats in guinea pigs. Cross two heterozygous guinea pigs to see the possibilities of their offspring. Parents = Bb x Bb B b Genotype: 25 % BB 50 % Bb 25 % bb BB Bb B Bb bb 1 Phenotype: 75 % black 25% brown b Oompa Lommpa Crosses 3. Oompa Loompas generally have gray faces (G) which is dominant. The recessive allele results in an orange face (g). Two heterozygous Oompa’s mate and have babies. What are the possible outcomes? Parents = Gg x Gg G g Genotype: 25% GG 50% Gg 25% gg Gg G Phenotype: 75% Gray faced GG 25% orange faced gg g Gg 4. A gray faced OOmpa (homozygous) is married to an orange faced Oompa. They have 8 children. How many of the children have gray faces? 8 How many have orange faces? 0 GG X gg 5. 100% Gg 100% Gray faced Otis Oompa has an orange face and is married to Ona Oompa who has a gray face. They have 60 Oompa children, 30 of their children have orange faces. What is Ona and Otis genotype? ____gg X Gg 50% Gg 50% gg Otis: gg Ona: Gg Let’s try something a little harder. 6. In Squid people, the allele for light blue skin (B) is dominant over the green (b) allele. Everyone in Squidward’s family has light blue skin. His family brags they are ‘purebred’ line. He recently married a nice girl with light green skin. What are the possible offspring outcomes? Would the children still be considered ‘purebred’? Why or why not? Light blue = BB (pure) light green = bb 100? Light blue skin (Bb) Children are not purebred (BB) 7. In goats, a recessive gene causes the goats to ‘faint’ when they are startled. A farmer breeds two goats (that have never fainted) and their first offspring faints two days after its birth. What must the parent’s genotypes have been? _____ x ______ Faint = recessive = ff Not fainting – FF or Ff Cross two non fainting goats: offspring F1faints so: Parents Ff X Ff generates 25% fainting goats ff 8. Two short-haired guinea pigs are mated several times. Out of 100 offspring, 25 of them have long hair. What are the probably genotypes of the parents? Ss x Ss 25% long hair (recessive) so parents must be both hybrids (Ss) for short hair 9. A tall plant of unknown genotype is test-crossed. Of the offspring, 869 are dwarf and 912 are tall. What is the genotype of the unknown parent? Tt Test cross: homoz. Dom. Or recessive X hybrid If dwarf is recessive tt, then, tall plant must be the hybrid, Tt 2 10. In humans, tongue rolling is a dominant trait. Those with the recessive condition cannot roll their tongues. Bob can roll his tongue, but his mother could not. He is married to Sally, who cannot roll her tongue. What is the probability that their first born child will not be able to roll his tongue? __50% tt________ Bob Can roll tongue is recessive, Tt (hybrid because his mother was recessive, tt) Sally can’t roll tongue, tt 11. If out of 100 offspring, 52 have red eyes and 48 have brown eyes. What are the probably genotype and phenotype of the parents? 52% red eyes 48% brown eyes Offspring is ~ 50% red 50% brown If red is recessive = rr If brown is dominant = R_ And Offspring is ~ 50% red 50% brown Offspring is ~ 50% red 50% brown Then both parents must be heterozygous Rr 12. A round (R) seed plant is dominant to a wrinkled (r) seed plant. What parental genotypes will produce offspring that are 50% homozygous dominant and 50% heterozygous? Round = dominant = R_ Wrinkled= recessive = rr Genotypes of parents if offspring is 50% RR : 50% Rr: One parent must be homozygous dominant and the other must be heterozygous Section C: Dihybrid Crosses 1. In guinea pigs, short hair is dominant to long hair and black eyes are dominant to red eyes. A male guinea pig that is homozygous dominant for both traits is crossed with a female who has long hair and red eyes. Complete a punnett square and determine the genotypic and phenotypic ratios. S= short hair s= long hair B = black eyes b= red eyes Parents: SSBB x ssbb Possible gametes: SB SB SB SB x sb sb sb sb SB sb SsBb SsBb SsBb SsBb SB SB SB Genotypes: 100% SsBb 3 sb SsBb SsBb SsBb SsBb Phenotypes: 100% short hair, black eyes sb sb SsBb SsBb SsBb SsBb SsBb SsBb SsBb SsBb You will need a separate sheet of paper for these problems. 2. In fruit flies, the allele for normal size wings (H) is dominant over the allele for vestigial wings (h). The allele for normal size eyes (E) is dominant over the allele for small eyes (e). The genes for wing size and eye size are on different chromosomes. Predict the results of a cross between a heterozygous individual for both traits and an individual who is homozygous recessive for both traits. Complete a punnett square and determine the genotypic and phenotypic ratios. Normal size wings = H Vestigial wings = h Normal size eyes = E Small size eyes = e HhEe x hhee HhEe possible gametes: HE, He, hE, he Hhee possible gametes: He, He, he, he HE He hE he He He he he GENOTYPIC RATIO: 2 HHEe : 2 HHee : 4 HhEe : 5 Hhee : 2 hhee : 1hhEe PHENOTYPIC RATIO: 6 normal wings, normal size eyes: 7 normal wings, small size eyes : 1 small wings, normal size eyes : 2 small wings, small size eyes 3. In racing horses, black hair (B) is dominant to chestnut hair (b) and a trotting gait (G) is dominant to a pacing gait (g). Cross two horses that are heterozygous for both traits. Complete a punnett square and determine the genotypic and phenotypic ratios. Black hair = B Chestnut hair = b Trotting gait = G Pacing gait = g 4 BbGg X BbGg Possible gametes BbGg: BG, Bg, bG, bg BbGg: same BG Bg bG bg BG Bg bG bg GENOTYPIC RATIO: 1 BBGG : 2BBGg : 2 BbGG : 2 Bbgg : 4 BbGg : 2 bbGg : 1bbgg : 1 bbGG : 1 BBgg PHENOTYPIC RATIO: 9 black hair, trotting gait : 3 black hair, pacing gait : 3 chestnut hair, trotting gait : 1 chestnut hair, pacing gait 4. Starting with a P generation with the following genotypes (AABB x aabb). Based on classical Mendelian inheritance, what is the expected phenotypic ration observed among the F2 generation? a. 9:3:3:1 b.1:2:1 c. 3:1 d. 1:1 e. 1:1:1:1 AABB x aabb : Cross between two homozygous parents with contrasting traits: F1 is 100% AaBb F2: AaBb X AaBb 5. If AaBb is crossed with aabb, what proportion of the offspring would be expected to be aabb? a. 9/16 b. 2/16 c. 4/16 d. 1/16 6. If the offspring of a cross show a 9:3:3:1 ration, the parents of the cross have the genotypes __. a. AaBb x AaBb b. AaBb x aaBB c. aaBb x aabb d. aaBb x Aabb Section D: Beyond Mendel – Patterns of Inheritance Codominance – 1. Crow’s (the black bird) feet can have orange markings or have brown markings. When a crow with orange (O) markings is mated with a crow with brown (B) markings, the resulting phenotype is both orange and brown spots on the feet. Cross an orange footed crow with a crow that has orange and brown spots on its feet. What are the genotypic and phenotypic ratios of the offspring? O O Orange markings = OO BO Brown markings = BB BO B Orange and brown markings = BO OO OO O OO X BO Genotypic ratio: 2 BO : 2 OO Phenotypic ratio: 2 orange markings : 2 black and orange markings 5 2. You are exploring the jungle and find a new species of plant. Some of the flowers are red and some are yellow. If the inheritance pattern is codominant, what would you expect the heterozygous phenotype to be? Red and yellow flowers Incomplete Dominance – 3. Rats can have a variety of fur color. When a black (B) furred rat is crossed with a white (W) furred rat, the resulting phenotype is gray fur. Cross a gray furred rat with a black furred rat. What are the phenotypic ratios of the offspring? Black fur = BB White fur = WW Gray fur = BW BW X BB B BB BB BW BW B B W Genotypic ratio: 2 BB : 2 BW Phenotypic ratio: 2 BLACK FUR : 2 GRAY FUR 4. If the red and yellow alleles in the mystery jungle plant above showed incomplete dominance, what would you expect the heterozygous phenotype to be? ORANGE Sex-Linked – 5. In fruit flies, eye color is sex-linked. Red is dominant to white. Cross a white eyed female X r X r with a red-eyed male X R Y. What are the genotypes and phenotypes of the offspring? R= RED Xr Xr r= white XRXr R r XrXr X XRY XXR X XrY XrY Y Genotypic ratio: 2 X R X r : 2 X r Y Phenotypic ratio: 2 red eye flies : 2 white eye flies 6. In humans, hemophilia is a recessive sex linked trait. Show the cross of a man who has hemophilia with a woman who is a Xh carrier. What is the probability that their children will have the disease? XH Hemophiliac man = X h Y h h X X Xh XHXh Carrier woman = XHXh XHY h h XhY h There is a 50% change that they’ll have a hemophiliac child (X X , X Y ) Y 7. A woman who is a carrier for hemophilia marries a normal man. If they have a daughter, what are the possible phenotypes? If they have a son, what are the possible phenotypes? 50% chance hemophiliac daughter, 50% chance hemophiliac boys Carrier woman = X H X h Xh Normal man = X H Y XH XhXh X HXX H XHY Y H XhY 6 8. Coat color in cats is a codominant trait and is also located on the X chromosome. Cats can be black, yellow or calico. A calico cat has black and yellow splotches. a. When a female calico cat is mated with a male black cat, what are the phenotypes of the offspring? Female calico = black and yellow = X B X Y Male black cat = X B Y X Phenotypes offspring: 1 black female: 1 calico female : 1 black male : 1 yellow male b. 9. B XY XBXY X BXX B XBY Y B XYY Where any of the offspring calico? yes If so, what gender was it? FEMALE A girl inherited colorblindness, which is a sex-linked recessive trait. It is probable, therefore, that __. a. Both parents had one colorblind gene. b. Only one parent had the colorblind gene. c. The gene in the mother guaranteed the girl having the trait. d. The father did not have the recessive gene. 10. A colorblind man marries a woman who is neither colorblind nor a carrier of the trait. Which statement best describes their probable offspring? a. All the boys will be colorblind. c. All the girls will be carriers but not have the disease. b. All the girls will be colorblind. d. Half the boys will carry the gene for colorblindness. Multiple Alleles – 11. In the ABO blood type system, the A and B are codominant and the o allele is recessive. A man with type AB blood marries a woman with type AB blood. If they have children, what are the possible phenotypes? 25% A blood 25% B blood 50% AB blood 0% o blood 12. A man with type AB blood is married to a woman with type o blood. They have two natural children and one adopted child. Jane has type A blood, Bobby has type B blood and Grace has type o blood. Which child was adopted? GRACE 13. A nurse at a hospital removed the wrist tags of three babies in the maternity ward. She needs to figure out which baby belongs to which parents, so she checks their blood types. Using the chart below, match the baby to its correct parents. Show the crosses to prove your choices. Parents Blood Types Baby Blood type Mr. Frisbee A Mrs. Frisbee B Dominic O Mr. Zimmerman O Kristen AB Mrs. Zimmerman O Tommy B Mr. Law AB Mrs. Law O DOMINIC: EITHER FRISBEE FAMILY OR ZIMMERMAN FAMILY 7 KRISTEN: FRISBEE FAMILY TOMMY: LAW FAMILY OR FRISBEE FAMILY Polygenic Inheritance – 14. In Snarlymonsters, the number of teeth is polygenic. The recessive condition (aabbcc) results in a toothless Snarlymonster, and the dominant condition (AABBCC) results in a Snarlymonster with 6 teeth. There are 5 other possible variations. How many teeth would a Snarlymonster with AaBbCc have? Lethal Genes – 15. In mice, the spinning behavior is caused by a dominant gene that affects the mouse's equilibrium. This gene is lethal if two alleles are present. Two "spinning mice" are mated together. Show the cross. What are the phenotypes of the offspring and in what proportion? I 1 II 1 1 2 2 3 3 2 4 4 5 6 7 8 5 Section E: Pedigrees III 1. 2. 3. 4. 5. There are no carriers for Huntington’s Disease- you either have it or you don’t. With this in mind, is Huntington’s disease caused by a dominant or recessive trait? ___________________ How many girls did II-1 and II-2 have? ___________ How many have the disease? ____________ If individual III-1 marries someone heterozygous for Huntington’s, then what’s the chance of having a child with Huntingtons? The pedigree to the right shows the passing on of colorblindness. What gender can be carriers? _________________ With this in mind, what kind of trait is colorblindness? _____________________ 8 III. 6. 7. Is this trait recessive or dominant? ________________ If individual IV-1 marries a carrier for colorblindness, then what’s the chance they will have a daughter who is colorblind? 9