While doing field work in Madagascar, you discover a new dragonfly species that has either red (R) or clear (r) wings. Initial crosses indicate that R is dominant to r. You perform three crosses using three different sets of red-winged parents with unknown genotype and observe the following data: Cross Phenotypes 72 red-winged, 24 clear-winged 4 red-winged 96 red-winged 1 2 3 What is the most likely genotype for Cross 1?
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- Individuals of genotype AaBb were mated to individuals of genotype aabb. One thousand offspring were counted, with the following results: 474 Aabb, 480 aaBb, 20 AaBb, and 26 aabb. What type of cross is it? Are these loci linked? What are the two parental classes and the two recombinant classes of offspring? What is the percentage of recombination between these two loci? How many map units apart are they?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In dogs, dark coat color phenotype (D) is dominant over albino (a) and short hair (S) is dominant over long hair (1). Assume that these phenotypes are caused by two independently assorting genes and write as much as you can, the genotypes of the parents in the cross below. D, a, S and I represent phenotypes Y indicates the presence of progeny with this phenotype , whereas N indicates the absence of progeny with this phenotype PARENTS PHENOTYPES TYPES of PROGENY PHENOTYPE D,S D,I a,S a,l D,S X a,l Y Y
- In the video game Animal Crossing: New Horizons, flowering breeding is based in genetics. Each flower's color is determined by the genotype at three or four unlinked genes: R, Y, W, and S. The genotype of the elusive blue rose is RR YY ww ss. In the game, one way to get a blue rose is to cross two roses with the Rr Yy Ww ss genotype. A) What types of gametes and in what proportions will a Rr Yy Ww ss rose produce? B) In a cross Rr Yy Ww ss x Rr Yy Ww ss what are the possible offspring genotypes and at what frequency will they each appear? Show your work. C) What proportion of the offspring of the cross will be blue roses?Fruit fly body color is wild type (meaning normal) Gray = B+ and black = b Fruit fly wing type is wild type normal wings = vg+ and vg = vestigial A fly which was heterozygous for both traits was crossed with a fly that was recessive for both traits. Write the genotype and phenotype that would be expected from this cross. Write the genotype and phenotype percentages that would be expected from this cross.In dogs, dark coat color phenotype (D) is dominant over albino (a) and short hair (S) is dominant over long hair (I). Assume that these phenotypes are caused by two independently assorting genes and write as much as you can, the genotypes of the parents in the cross below. D, a, S and I represent phenotypes Y indicates the presence of progeny with this phenotype, whereas N indicates the absence of progeny with this phenotype PARENTS PHENOTYPES TYPES of PROGENY PHENOTYPE D,S D,I a,S a,l D,S X a,l Y Y N
- In dogs, dark coat color is dominant over albino, andshort hair is dominant over long hair. Assume that theseeffects are caused by two independently assorting genes.Seven crosses were done as shown below, in which D andA stand for the dark and albino phenotypes, respectively,and S and L stand for the short-hair and long-hairphenotypes.Number of progenyParental phenotypes D, S D, L A, S A, La. D, S × D, S 88 31 29 12b. D, S × D, L 19 18 0 0c. D, S × A, S 21 0 20 0d. A, S × A, S 0 0 29 9e. D, L × D, L 0 31 0 11f. D, S × D, S 45 16 0 0g. D, S × D, L 31 30 10 10Write the genotypes of the parents in each cross. Use thesymbols C and c for the dark and albino coat-color allelesand the symbols H and h for the short-hair and long-hairalleles, respectively. Assume parents are homozygousunless there is evidence otherwise.Two pure-breeding lines of petunia plants are crossed. Line 1 plants grow to a height of 54 cm, and Line 2 plants grow to a height of 18 cm. Petunia plant height is controlled by three genes, A, B and C. Line 1 has the genotype A₁A₁B₁B₁C₁C₁, and line 2 has the genotype A2A2B₂B₂C₂C₂. Assume that genotype alone determines plant height under ideal growth conditions and that the alleles of the three genes are additive. If the F1 plants are self crossed, what is the expected proportion of F2 plants with the genotype A₁A₁B₁B₁C₁C₁ 1/8 1/32 1/16 1/4 1/64You perform a cross between a parent with the genotype WWiiNNttEErr and another parent that is wwllnnTTeerr. All genes are unlinked except for W and I which are 22 mu apart. You take an F1 from this cross and cross it with an individual that is wwiiNntteerr. a) What is the probability that this final cross yields an offspring that is wwiinntteerr? b) What is the probability that this final cross yields an offspring that is NNTT or Nntt. (You can ignore all of the other genes for this question.)
- In a species of the cat family, eye color can be gray, blue, green, or brown, and each trait is true breeding. In separate crosses involving homozygous parents, the following data were obtained: Cross P1 F1 F2 A green * gray all green 3/4 green: 1/4 gray B green * brown all green 3/4 green: 1/4 brown C gray * brown all green 9/16 green: 3/16 brown 3/16 gray: 1/16 blue (a) Analyze the data. How many genes are involved? Define gene symbols and indicate which genotypes yield each phenotype. (b) In a cross between a gray-eyed cat and one of unknown genotype and phenotype, the F1 generation was not observed. However, the F2 resulted in the same F2 ratio as in cross C. Determine the genotypes and phenotypes of the unknown P1 and F1 cats.In pineapples, leaves may be spiny, spiny-tipped or piping. Pineapples of different phenotypes were used in the following crosses: Pl phenotype piping X spiny P1 genotype Fl phenotype 100% piping F1 genotype Crossing the Fls produced: F2 phenotypes F2 genotypes ww gm 95 piping 25 spiny tipped 8 spiny Using the letters A and B, give the COMPLETE genotypes of ALL the individuals. SOLUTION Get the total of the F2 generationAn ebony strain of flies was discovered to be sensitive to carbon dioxide. Crossing a female sensitive strain with male resistant strain gave all sensitive offspring. The offspring of an F1 female crossed with a resistant male were all sensitive. Using the following key to your illustrations using shapes, make a reciprocal cross up to the F2 generation. Put your illustrations in the space provided below. Label the phenotypes of all individuals in the reciprocal cross. Adjust spacing, if necessary. Make sure that the complete cross(es) can fit the same page. Big blue circle - male cytoplasm Big pink circle - female cytoplasm Small half-blue-half-pink circle - F1 nucleus Small blue circle - male nucleus Small pink circle - female nucleus give a diagram please