Answer Happy

(From these crosses you conclude that the mutations in all threelines are recessive to the wild type.)
PART A
PART B
PART C
You study color variants of Arabidopsis hypotheticus, a plant with red flowers. You have obtained three pure-breeding mutant lines, all named for their place of origin. Two lines have white flowers (Aberdeen White and Victoria White), and one has yellow flowers (Kansas Yellow). You begin your analysis by crossing each line with wild-type plants and selfing the F₁ to produce an F2 generation. fo Cross Aberdeen White X Victoria White faf X Kansas Yellow X wild type wild type wild type F₁ phenotype wild type wild type wild type F₂ phenotypes and ratios 3 wild type : 1 white 3 wild type : 1 white *) 3 wild type : 1 yellow
Part A - Crossing lines with the same recessive phenotype You continue your analysis by crossing the Aberdeen White and Victoria White lines. This time you count the actual numbers of progeny in the two F2 phenotypic classes. F2₂ phenotypes and numbers dominant epistasis independent assortment recessive epistasis complementary gene action You also self several wild-type plants from the F₂ and determine that some of them are pure-breeding. What can you conclude from these results? For each statement, choose the correct response by selecting only from the three answer choices to the left of that statement. Drag the correct answer to the right of each statement. ► View Available Hint(s) dominant epistasis 13:3 9:7 3:1 false true Cross cannot be recessive epistasis determined Aberdeen White x Victoria White 1. Aberdeen White and Victoria White are alleles of different loci. F₁ phenotype 2. The pure-breeding wild-type progeny in the F₂ are the result of 3. The F₂ ratio is approximately 4. The F₂ ratio results from wild type Group 1 Group 2 Group 3 181 wild type, 139 white Group 4 Reset Help
Part B - Crossing lines with dissimilar recessive phenotypes You continue your analysis by crossing the Aberdeen White and Kansas Yellow lines. Again, you count the actual numbers of progeny in the two F2 phenotypic classes. F₂ phenotypes and numbers true failure to complement 3:1 (from a 1:2:1 genotype ratio) recessive epistasis false What can you conclude from this cross? For each statement, choose the correct response by selecting only from the three answer choices to the left of that statement. Drag the correct answer to the right of each statement. ► View Available Hint(s) recessive epistasis 13:3 (from a 9:3:3:1 genotype ratio) complete dominance Aberdeen White x Kansas Yellow cannot be determined dominant epistasis Cross 12:4 (from a 9:3:3:1 genotype ratio) dominant epistasis F₁ phenotype fa yellow 1. Aberdeen White and Kansas Yellow are alleles of different loci. 2. The lack of wild-type progeny in the F₂ is the result of 3. The F₂ ratio is approximately 121 yellow, 41 white 4. The F₂ ratio results from Reset Group 1 Group 2 Group 3 Group 4 Help
Part C- A second cross between lines with dissimilar phenotypes You complete your analysis by crossing the Victoria White and Kansas Yellow lines. Again, you count the actual numbers of progeny in the F₂ phenotypic classes. F₂ phenotypes and numbers true dominant epistasis 12:3:1 dominnat epistasis As before, you self several wild-type plants from the F2 and again determine that some of them are pure-breeding. What can you conclude from these results? For each statement, choose the correct response by selecting only from the three answer choices to the left of that statement. Drag the correct answer to the right of each statement. ▶ View Available Hint(s) false recessive epistasis 1:2:1 recessive epistasis Victoria White x Kansas Yellow cannot be determined independent assortment Cross 9:3:4 duplicate gene action F₁ phenotype wild type 1. Victoria White and Kansas Yellow are alleles of different loci. 2. The pure-breeding wild-type progeny in the F₂ are the result of 3. The F₂ ratio is approximately 4. The F₂ ratio results from 89 wild type, 39 white, 29 yellow P Pearson Reset Help Group 1 Group 2 Group 3 Group 4