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- 3. For the following crosses, describe the genotype and phenotype frequencies of the offspring. I recommend using Punnett squares for each. Offspring phenotypes and frequencies А - 50% О -50% Parental Offspring genotypes and frequencies Parental genotypes phenotypes LAj x i Ai – 50% ii – 50% A and O Bị x Ai BIB x Ai AA x IBIB Bị x Bi AB x i3. If an individual with genotype Yy Ww Pp ee Bb is crossed with an individual with the same genotype, what proportion of offspring will have the dominant phenotype for the four heterozygous loci? answers: 81/256 81/128 27/256 3/161. The petals of the blue-eyed Mary (Collinsia parviflora) are normally blue. There are color variants of white and pink. Two pure breeding lines, one with pink petals and another with white petals were collected and the following crosses were made, with results shown in the table below: parents F1 F2 blue x white blue 101 blue, 33 white blue x pink blue 192 blue, 63 pink pink x white blue 272 blue, 121 white, 89 pink A. Define the allele symbols that can be used and show the genotype of parents, F1 and F2. B. A cross between a certain blue F2 plant and a certain white F2 plant gave progeny of which 3/8 were blue, 1/8 were pink, and ½ were white. What must the genotypes of these two F2 plants have been?
- 4. Attached earlobes are dominant over free hanging earlobes. Complete the Punnett Square for the following individuals: Mom-Bb and Dad=bb Provide the genotype and phenotype ratios using the Punnett square given below. Genotypes 5. When two alleles are observed independently in a phenotype is called multiple alleles. Human blood group ABO is a good example of multiple alleles. The three alleles are: IA, IB, i. Mother's blood type is AB (IAI³) and father's blood type is B (I³i). Provide the genotype and phenotype ratios using the Punnett square given below. Genotypes Phenotypes Genotypes Phenotypes 6. When the allele of a trait is found on X or Y chromosome, it becomes sex-linked. The results differ from males to females. One such example is of red green colorblindness. The mother is a carrier (X© Xº ), and the father is colorblind (X© Y). Provide the genotype and phenotype ratios using the Punnett square given below. PhenotypesI. Chi-Square Test Phenotype Class Observed Frequency (0) Seed Expected Frequency (E) Deviation d2 d?/E (d=0-E) Corn Tall-normal 26 9 17 289 3.01 (Corn) Rice Tall-wrinkled 26 3 23 529 5.51 (Rice) Mongo Dwarf-normal 25 3 22 256 2.66 (Mongo) Squash Dwarf- 19 3 116 484 5.04 wrinkled (Squash) 16.22 Degree of freedom - 4-1-3 EL X' = S lau -8) 23 G,089 18 %3D X'= 338 Exercise No. 4- Gene Segregation and Interaction 4 What can you conclude based on the value of the computed Chi-square? How can you relate the two principles of Mendel to Chi-Square Values?Once upon a time there was a population of hippies living in an isolated commune called Gone. They had been there for generations, and all members were homozygous for a gene that gave them a preference for crafting necklaces out of green beads, so that every individual had the genotype GG at this locus. In another isolated commune, called Yonder, was a population of hippies that had a genetic propensity for making necklaces out of yellow beads, and everyone was homozygous for a different allele at the same genetic locus; their genotype was YY. Last year (prior to our current social isolating) everyone repaired their VW buses, and a giant gathering in celebration of the 50th anniversary was held at the Oregon Country Fair. The next few questions will ask you about these two populations that came together at the fair.
- Using the following information perform following crosses using the forked line and predict all possible phenotypes types of the offspring. List the number and description of the phenotypes of the offspring. 1. aa BB Cc Dd Ee X aa bb Cc Dd EE 2. Ff Rr Gg dd Bb X Ff RR Gg Dd Bb 3. Dd gg JJ bb Hh Ee X DD GG Jj Bb Hh EePractice: 1.) Heterozygous A crosses with Heterozygous A. Calculate the probability of each phenotype being passed down to the offspring. TOTALS: (percent and ratio) A: B: AB: O: 2.) Type O breeds with Type AB TOTALS: (percent and ratio) A: B: AB: O: 3.) Woman has type B (exact genotype unknown). Man has type O. What are the possible blood types of their kids? Is this type possible: A: yes/ no? B: yes/ no? AB: yes/ no? O: yes/ no? 3B.) If the woman has a baby with type AB blood, could the man above be the father? Explain you answer.1. A selection experiment was carried out on a population of field mice in the lab had a mean hematocrit of 65. The mice used to create the next generation had a mean hematocrit of 50. Once the offspring from that experiment reached adulthood, the experimenters measure a mean hematocrit of 58. Answer the following questions using this information a). What is the narrow sense heritability for hematocrit in these mice? Show all work b).Using those results, what would be the predicted hematocrit in the offspring population after another generation of selection where only mice with a mean hematocrit 10 points below the parental mean were allowed to breed. Show all work c). Would you expect the narrow sense heritability for a wild population of field mice in Iqaluit to be the same ? Explain.
- a. 1 dominant allele will contribute 120/10 = 12 cm to the base height of the plant.b. The height of the parent plant 1 Genotype of the parent plant 1 – D1D1D2D2D3D3d4d4d5d5 The height of the parent plant 2 Genotype of the parent plant 2 – d1d1d2d2d3d3D4D4D5D5Contributing alleles – D4D4D5D5. The height of the plant without any contributing alleles would be 80 cm. The plant with genotype d1d1d2d2d3d3D4D4D5D5 has 4 contributing allele each of which contributes 12 cm to the base. Hence, the height of the plant with genotype d1d1d2d2d3d3D4D4D5D5 would be 80 + 12 + 12 + 12 + 12 = 128 cm. c. Parents – D1D1D2D2D3D3d4d4d5d5 × d1d1d2d2d3d3D4D4D5D5 Gametes – D1D2D3d4d5 × d1d2d3D4D5 F1 generation – D1d1D2d2D3d3D4d4D5d5 The height of the plants of F1 generation = 80 + 12 + 12 + 12 + 12 + 12 = 140 cm Hence, Genotype of the F1 = D1d1D2d2D3d3D4d4D5d5 Phenotype of…4 D 5 D U 6 D U 7D Genotypes are TT, Tt, and tt. Phenotypes are Taster or Non-taster. Genotype Phenotype 1 Non taster taster taster 131 %3D 2. 3.2. Show the cross and provide full justification in each case: a) An F1 x F1 self-fertilization gives a 9:7 phenotypic ratio in the F2. What phenotypic ratio would you expect if you test-crossed the F1? b) An F1 x F1 self-fertilization gives a 9:3:4 phenotypic ratio in the F2. What phenotypic ratio would you expect if you test-crossed the F1? c) An F1 x F1 self-fertilization gives a 15:1 phenotypic ratio in the F2. What phenotypic ratio would you expect if you test-crossed the F1?