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- . A geneticist mapping the genes A, B, C, D, and E makestwo 3-point testcrosses. The first cross of pure lines isA/A ⋅ B/B ⋅ C/C ⋅ D/D ⋅ E/E × a/a ⋅ b/b ⋅ C/C ⋅ d/d ⋅ E/EThe geneticist crosses the F1 with a recessive tester andclassifies the progeny by the gametic contribution ofthe F1:A ⋅ B ⋅ C ⋅ D ⋅ E 316a ⋅ b ⋅ C ⋅ d ⋅ E 314A ⋅ B ⋅ C ⋅ d ⋅ E 31a ⋅ b ⋅ C ⋅ D ⋅ E 39A ⋅ b ⋅ C ⋅ d ⋅ E 130a ⋅ B ⋅ C ⋅ D ⋅ E 140A ⋅ b ⋅ C ⋅ D ⋅ E 17a ⋅ B ⋅ C ⋅ d ⋅ E 131000The second cross of pure lines is A/A • B/B • C/C • D/D •E/E × a/a • B/B • c/c • D/D • e/e.The geneticist crosses the F1 from this cross with arecessive tester and obtainsA ⋅ B ⋅ C ⋅ D ⋅ E 243a ⋅ B ⋅ c ⋅ D ⋅ e 237A ⋅ B ⋅ c ⋅ D ⋅ e 62a ⋅ B ⋅ C ⋅ D ⋅ E 58A ⋅ B ⋅ C ⋅ D ⋅ e 155a ⋅ B ⋅ c ⋅ D ⋅ E 165a ⋅ B ⋅ C ⋅ D ⋅ e 46A ⋅ B ⋅ c ⋅ D ⋅ E 341000The geneticist also knows that genes D and E assortindependently.a. Draw a map of these genes, showing distances inmap units wherever possible.b. Is there any evidence of interference?9. Make a pedigree for each of the following situations. For each individual, write the individual's genotype (when possible) next to the individual's symbol (e.g. O xty, I Gg): a. Two parents do not have cystic fibrosis and they have a daughter with cystic fibrosis and a son who does not have cystic fibrosis. The daughter grows up and she mates with a male who does not have cystic fibrosis. Their only child is a boy and he has cystic fibrosis. b. A man with hemophilia mates with a female without hemophilia. They have one son and one daughter. The daughter has hemophilia and the son does not have hemophilia. The son grows up, and he marries and mates with a female. Their only child is a boy, and he has hemophilia.Purple flowers are dominant to white flowers. Identify the phenotypefor the following genotype Ff, FF, ff and determine if the genotype is heterozygous or homozygous. * For each row, you should select two columns. Purple flowers White flowers Heterozygous Homozygous Ff FF ff Brown eyes are dominant to blue eyes. Identify the phenotype for the following genotype BB, Bb, bb and determine if the genotype is heterozygous or homozygous. * 口 ロ口
- FAlpQLSfiOhfAvlhxzCSiUll_6rt-nU5b0WI73UmWOxkOw8OCwk01ng/formResponse B 1 2 Bb x Bb b 4 The fur in both parents in this cross is * 1 B B Bb x Bb b 3 4 brown black O homozygous dominant homozygous recessive 3. 近Hi, I'm having trouble with my study guide for my upcoming genetics exam. If someone could please help with work shown and an explanation it would help so much! Thank you!! 2a. The pedigree below represents inheritance of rare condition. What pattern of inheritance is most consistent with the data? Assign alleles to all individuals to support your answer. If an allele is unknown, assign it a ? symbol. NOTE: Individuals whose phenotype or genotype cannot be determined are assumed to be unaffected and homozygous, unless otherwise indicated. 2b. In addition to the alleles you’ve indicated, describe 2 overall features of the pedigree that make it consistent with your chosen form of inheritance. 2c. Based on your mode of inheritance, what is the probability that the child of couple IV-4 x IV-5 will be affected? Show your work. attached is the pedigreeA corn geneticist wants to obtain a corn plant that hasthe three dominant phenotypes: anthocyanin (A), longtassels (L), and dwarf plant (D). In her collection ofpure lines, the only lines that bear these alleles are AALL dd and aa ll DD. She also has the fully recessive lineaa ll dd. She decides to intercross the first two and testcross the resulting hybrid to obtain in the progeny aplant of the desired phenotype (which would have to beAa Ll Dd in this case). She knows that the three genesare linked in the order written, that the distance between the A/a and the L/l loci is 16 m.u., and that thedistance between the L/l and the D/d loci is 24 m.u.a. Draw a diagram of the chromosomes of the parents,the hybrid, and the tester.b. Draw a diagram of the crossover(s) necessary toproduce the desired genotype.c. What percentage of the testcross progeny will be ofthe phenotype that she needs?d. What assumptions did you make (if any)?
- 1. Use the below pedigree chart to answer the following questions about dimples. The Dimple gene controls whether a person has dimples or doesn't have dimples. Dimples is dominant to no dimples. Place the genotypes of each individual below its symbol. male female 7 10 11 male femake Dimples gene (D) Dimples is dominant to no dimples 12 13 14 A) How many family members have Dimples? B) What is the genotype of individual #3 and 4? C) Can either individual #8 or 9 be homozygous? C) Explain the family relationship that #12 has with # 2.1. Use the below pedigree chart to answer the following questions about dimples. The Dimple gene controls whether a person has dimples or doesn't have dimples. Dimples is dominant to no dimples. Place the genotypes of each individual below its symbol. male female 6 7 8 10 11 male female Dimples gene (D) Dimples is dominant to no dimples 12 13 14 May 2 ||: 59 A) How many family members have Dimples? B) What is the genotype of individual #3 and 4? C) Can either individual #8 or 9 be homozygous? C) Explain the family relationship that #12 has with # 2.CH 1-4 X с Maya S x Credib x app.wizer.me/learn/00E0AB wizer.me a Maya X A To-do 25% The table gives the genoty Assign x M Dashb x 50% Figurat X C In cows, brown (B) is codominant with white (W). The heterozygous phenotype is brown and white speckles. A farmer decided to cross a brown cow and a white cow in hopes of making all brown and white speckles. What percentage of the offspring will be brown with white speckles? Figurat X 75% Dashboard Incomp X d Interac X Enter class code Go 100%
- This pedigree (Pedigree #2) illustrates the inheritance of a simple Mendelian trait. If individuals III- 5 and III-6 have children, what are the chances that the children would have this disorder? 0% 1/4 2/3 O 1/6 ㅇㅁEach box represents a potential offspring. Notice that genotypically speaking, they are all the same. Theyare all heterozygous (genotype) which means their phenotype is dark coloration. So, there is 100% chanceof producing a dark eyed, heterozygous offspring. What if that offspring mated with a light eyedindividual? Can you make a Punnett Square for that?Write the Genotypes and Phenotypes of the parents on the left of the square. Complete the square, thenwrite the potential offspring’s genotypes and phenotypes on the right of the square. Parents: Offspring:Genotypes: phenotypes: What are the genotypes of the resulting offspring? What are the phenotypes of the resulting offspring?2. Figure 8.10 In pea plants, round seed shape (R) is dominant to wrinkled seed shape (r) and yellow peas (Y) are dominant to green peas (y). What are the possible genotypes and phenotypes for a cross between RrYY and rrYy pea plants? How many squares do you need to do a Punnett square analysis of this cross?