Use the following image to answer the question. If you performed the cross indicated to the F 2 generation, how many smooth, green peas would you expect to see in 100 offspring? YYRR yyrr P Generation а. 6 Ob. 19 c. 57
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- Compute the following probabilities for a selfing plant using Figure 6.2.9. The fraction of fourth-generation offspring with genotype AAUse the following image to answer the question. If you performed the cross indicated to the F 2 generation, how many smooth, yellow peas would you expect to see in 100 offspring? YYRR yyrr P Generation а. 6 b. 57 Ос. 19Use the image to observe the results of a cross between a tall pea plant and a short pea plant. What phenotypes and proportions will be produced for the two crosses? P generation Tall Tt Gametes Tt F₁ generation T Fertilization Tt t Tall tt Short Tt Tall tt tt Short Conclusion (a) Short (b) Genotypic ratio 1Tt:1tt Phenotypic ratio 1 tall: 1 short Tall F₁ progeny backcrossed to the short parent Answer Bank one-fourth tall and three-fourths short one-half tall and one-half short three-fourths tall and one-fourth short Tall F₁ progeny backcrossed to the tall parent all tall all shor
- > NEXT A BOOKMARK Question 23/28 According to the probability of the pea plant offspring in chart 2, what percentage of the offspring will be yellow in color? Use the diagrams below to answer the questions that follow. 23 Pea Plants Chart 1 A 5% Punnett Square B) 19% с) 30% Aa Aa D 25% Aa A a E) 75% Chart 2 AaBb x AaBb AB Ab aB ab Key AB AABB AABB Ab AABB AABB AaBb AA - Green BB - Smooth AAbb AaBb Aabb AABB AaBb aa - Yellow aB AaBb aaBB aaBb bb - Rough ab Aabb aaBb aabbUse the following information to answer the next question. In a breeding experiment, a pure plant with round seeds and green pods (RRGG) was crossed with another pure plant with wrinkled seeds and yellow pods (rrgg). The phenotypes of the F2 generation were: 1780 plants with round and green pods, 620 plants with round seeds and yellow pods, 590 plants with wrinkled seeds and green pods and 195 plants with wrinkled seeds and yellow pods. Which law is proved by the result of this experiment? Law of Segregation Law of Mutation Law of Independent Assortment Law of DominanceA researcher crosses two tall plants, however, one quarter (1/4) of the offspring from this crossing have short stems. Complete the punnett square. 6 b EBE Bb Besides Bb, what is the other possible genotype for a tall plant? What is the genotype of a short plant? Bb bb
- A snapdragon with pink petals, black anthers, and longstems was allowed to self-fertilize. From the resultingseeds, 650 adult plants were obtained. The phenotypesof these offspring are listed here.78 red long tan26 red short tan44 red long black15 red short black39 pink long tan13 pink short tan204 pink long black68 pink short black5 white long tan2 white short tan117 white long black39 white short blacka. Using P for one allele and p for the other, indicatehow flower color is inherited.b. What numbers of red : pink : white would havebeen expected among these 650 plants?c. How are anther color and stem length inherited?d. What was the genotype of the original plant?e. Do any of the three genes show independentassortment?f. For any genes that are linked, indicate the arrangements of the alleles on the homologous chromosomes in the original snapdragon, and estimate thedistance between the genesComplete the punnett square below for a cross between AA and aa A 21 a QUESTION 4 A Aa Complete the punnett square below for a cross between Aa and Aa. 03 Aa AaWhat would justify the following ratio appearing after phenotyping the outcome of a crossing trial: 8.9: 2.9: 3.2:1? Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a b C d Obviously this represents independent assortment based on crossing dihybrid heterozygotes. Obviously this represents gene linkage based on test crossing a dihybrid heterozygote. Obviously this represents the results of a trihybrid test cross. Obviously this represents independent assortment based on crossing monohybrid heterozygotes.
- 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…Give the genotype of the parents and determine the linkage map of the three genes bm (brown midrib), v (virescent seedling), and pr (purple aleurone) in maize from the results of the cross below. Genotypes of offspring Total and Frequency percentage V bm 230 467 pr + 237 42.1% + + bm 82 161 pr V 79 14.5% V 200 395 pr bm 195 35.6% pr bm 44 86 V + 42 7.8% +In a diploid plant species, an F1 with the genotype Gg Ll Tt is test-crossed to a pure-breeding recessive plant with the genotype gg ll tt. The offspring genotypes are listed in the table. Genotype Number Gg Ll Tt 621 Gg Ll tt 3 Gg ll Tt 64 Gg ll tt 109 gg Ll Tt 103 gg Ll tt 67 gg ll Tt 7 gg ll tt 626 1600 Calculate the recombination frequency between G and T pair of genes. A. 0.227 B. 0.139 C. 0.454 D. 0.233