5. Design a mating between a female with these three traits: purple (eye color), black (bodycolor), and curved (wing shape), and a wild type male.a. What are the phenotypes of their offspring?+ (wildtype)b. What are their genotypes?XW+ XW+XW+ Y9Female PRCSource Design ViewQFyFlies to be MatedPhenotypeFemale: +Male: +TotalMaleSource Design ViewResults for Cross # 3Observed Proportion5220.511349910210.48871.0000Female: PR, C, BL; N=18Parents of Cross #3Select to MateQ Select to MateOffspring of Cross #3Male N-499Female N 522Select to Male10 TotalMale N=1PhenotypeSelect to MateResults for Cross #3Observed Proportion10211.000010211.0000

Linkage is a phenomenon in which genes are placed on the same chromosome in close proximity to each…

5. Design a mating between a female with these three traits: purple (eye color), black (body color), and curved (wing shape), and a wild type male. a. What are the phenotypes of their offspring? + (wildtype) b. What are their genotypes? Xw+ Xw+ XW+ Y 9 Femme P. C. Sou Dein View Saty Flies to be Mated Results for Cross # 3 Phenotype Female: Male: + Total View Observed Proportion 522 0.5113 0.4887 1.0000 499 1021 Parents of Cross #3 Female: PR, C. B: N-1 Q Fem-522 a Select to M Offspring of Cross #3 10 Total Male - N-1 Phenotype QS to Ma Mae - 499 Results for Cross #3 Observed Proportion 1021 1.0000 1021 1.0000

5. Design a mating between a female with these three traits: purple (eye color), black (body color), and curved (wing shape), and a wild type male. a. What are the phenotypes of their offspring? + (wildtype) b. What are their genotypes? Xw+ Xw+ XW+ Y 9 Femme P. C. Sou Dein View Saty Flies to be Mated Results for Cross # 3 Phenotype Female: Male: + Total View Observed Proportion 522 0.5113 0.4887 1.0000 499 1021 Parents of Cross #3 Female: PR, C. B: N-1 Q Fem-522 a Select to M Offspring of Cross #3 10 Total Male - N-1 Phenotype QS to Ma Mae - 499 Results for Cross #3 Observed Proportion 1021 1.0000 1021 1.0000

Biology: The Dynamic Science (MindTap Course List)

4th Edition

ISBN:9781305389892

Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Publisher:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Chapter12: Mendel, Genes, And Inheritance

Section: Chapter Questions

Problem 4TYK: What genotypes, and in what frequencies, will be present in the offspring from the following...

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Pa × PP Dd 6/6 - Long beak Short beak 7. What percentage of the offspring will have a homozygous dominant genotype? 8. What will be the ratio of long beaks to short beaks in the offspring? LI x 11 9. What percentage of the offspring will White flowers Purple flowers have purple flowers as their phenotype? 10. What percentage of the offspring will have a heterozygous genotype? pp х x PP 11. What percentage of the offspring will Dark eyes Dark eyes have a heterozygous genotype?
2. Cross a homozygous two horned zork with a heterozygous two horned.zork. A. What are the genotypes of the possible offspring? H Gendrere ны 3. Cross alheterozygous green haired zork with a heterozygous green haired zork. A. What are the genotypes and phenotypes of the possible offspring? ✓homozygous 4. Cross a green lipped zork with a heterozygous purple lipped zork.) A. What are the number of phenotypes and genotypes of the offspring? Hint: Count what is in the boxes! 5. Tork, who is homozygous for tall meets Vorkina, who is short. A. What are the phenotypes and genotypes if they were to have offspring?
2. In the following types of matings, the phenotypes of the parents are listed together with the frequencies of phenotypes occurring among their offspring. Indicate the genotype of each parent (you may need to use testcrossesl). Parents Offspring I B:0 을 A: 을B 을 AB : B: A: 0 를 AB : 를 A a. Вx В Ь. Ох АВ Ox AB Bx A d. Bx A C. TV Sex Lin ko
Im pP x PP 6/6 wa s hatarosygous genotypa? Dark eyes Dark eyes 11. What percentage of the offspring will have a heterozygous genotype? 12. What will the ratio of flies with dark eyes to white eyes be? Dd x x Dd 13. What percentage of the offspring will be short? Tall Tall 14. What percentage of the offspring will have a homozygous dominant genotype? TT х Tt 15. What percentage of the offspring will Long beak Long beak have long beaks?
I 2. In frost moths, two alleles of one gene determine the character difference of spotted versus striped wings and two alleles of a separate, independent gene determine the character difference of orange wing background versus white wing background. The results for four matings of moth phenotypes are shown below. [Yes - it's possible for the same phenotypes to have different genotypes]. Mating Parental Phenotypes 1 2 3 4 spotted/white x striped/orange spotted/white x striped/orange spotted/orange x striped/ white spotted/orange x striped/orange Spotted/ orange 122 812 422 623 Number of progeny Spotted/ Striped / white orange 131 128 0 430 0 637 a. Determine which alleles are dominant and assign a letter to each. Striped / white 133 0 0 0 b. What are the genotypes of each parent in each cross? If more than one genotype is possible, list the alternative possibilities.
What genotypes, and in what frequencies, will be present in the offspring from the following matings? a. AABBaaBB b. AaBbaabb c. AaBBAABb d. AaBbAaBb
7. Here are the results of a crossing trial with fruit flies (D. melanogaster) Parents: Ebony bodied, white-eyed 9 fly x Normal-bodied, red-eyed a fly • Both parents were from true-breeding (homozygous) lines • Ebony body means the body colour is black instead of orange F1: Males All white-eyed, normal bodied Females Red eyed, normal bodied F2: (200 offspring in total produced) Male Females Normal body, red eye Normal body, white eye Ebony body, red eye Ebony body, white eye Total 38 36 37 39 12 11 13 14 100 100 Explain the pattern of inheritance of each trait body colour and eye colour Write the genotype of the parents, F1 offspring and F2 offspring. (ii)
Part 1 Monohybrid crosses 1. The following lists represent genotypes at imaginary genetic loci. In each case there may be one or more correct answers – list ALL correct answers a. Homozygous dominant Mm NN oo pp Qq RR b. Homozygous recessive Mm NN oo pp Qq RR c. Genotypes in which dominant allele must show Mm NN oo pp Qq RR d. Genotypes in which recessive allele must show Mm NN oo pp Qq RR
III. Dihybrid crosses Two loci with two autosomal recessive traits Gene symbols a apterous wing size A normal wing size s sepia eye color S red eye color Be careful, S and s look alike when hand written The parents were apterous, red eyed color female and normal wing size, sepia eyed coloredmale (figure out the genotypes) Diagram of P1 cross: You are crossing AaSs with AaSs. Diagram of F1 cross: Give expected F2 phenotypic ratio: Give expected F2 genotypic ratio: IV. Other dihybrid cross Gene symbols v vestigial wing size V normal wing size s sepia eye color S red eye color The parents were vestigial, red eyed color female and normal wing size, sepia eyed color male…
3. In the X linked fruit fly cross , why were the adult flies in the XRXr x XRY vial removed? 1. Adult fruit flies are known to eat their larvae and this would throw off the count 2. The goal was to count the phenotypic ratio of the offspring; therefore the parental generation must be removed. 3. The adult flies would eat all the food and the offspring would die before a count could be made
15. B. You set up a cross between a male fruit fly with lobe eyes and a true breeding female fruit fly with reduced bristles and disrupted wings. In the F1 offspring you see the all the flies have WT bristles and WT wings. However, half the flies have WT eyes and half the flies have lobe eyes. You take 20 of the F1 female flies with lobe eyes and mate them to 20 true breeding male flies with reduced bristles and disrupted wings and observe the following offspring: 164 lobe eyes and reduced bristles 137 lobe eyes and disrupted wings 638 reduced bristles and disrupted wings 152 reduced bristles 40 wildtype 175 disrupted wings 659 lobe eyes 35 lobe eyes reduced bristles and disrupted wings Calculate the interference for this region.
}} >> 用行书 1 6 88356 11 新日員 13 38 19 2 ? 14 老式 20 3 DE 15 68 21 O a. a normal male karyotype O b. a male who is missing a y chromosome O c. a female who is missing a y chromosome O d. a normal female karyotype 9 22 18 16 8 8等 1盒 11 x 17 in Y 5 Si 12 18