In kangaroos, pouched kangaroos (P) are dc bushy tail of unknown genotype so you dec kangaroos have bushy tails and 50% of the PPBB PPBb PPbb PpBB PpBb Ppbb ppbb
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- Figure 8.10 In pea plants, purple flowers (P) are dominant to white (p), and yellow peas (Y) are dominant to green (y). What are the possible genotypes and phenotypes for a cross between PpYY and ppYy pea plants? How many squares would you need to complete a Punnett square analysis of this cross?Imagine that you are performing a cross involving seed color in garden pea plants. What traits would you expect to observe in the F1 offspring if you cross true-breeding parents with green seeds and yellow seeds? Yellow seed color is dominant over green. a. only yellow-green seeds b. only yellow seeds c. 1:1 yellow seeds:green seeds d. 1:3 green seeds:yellow seedsIn the javelina an allele for albino (a) is recessive to the wild type color (A). At another locus, curly tails (C) is dominant to straight tails (c). You cross a homozygous wild type color, curly tailed javelina with an albino, straight tailed javelina. What is the genotype and phenotype of the F1 offspring? If you intercross the F1, what is the expected phenotypic ratio for the F2? In your F2 generation you get 270 wild type color, curly tailed javelina; 100 albino, straight tailed javelina; and 10 albino, curly tailed javelina. Use a Chi-Square test to determine if these results fit your hypothesis (b) What is your new hypothesis to explain the relationship between the genes that code for color and tail shape in javelina? Do those results fit your new hypothesis? (Test again using the Chi-Square Goodness of Fit Test)
- In kangaroos, pouched kangaroos (P) are dominant over pouchless kangaroos (p) and bushy tails (B) are dominant over skinny tails (b). You perform the cross below and observe the progeny (these kangaroos are very prolific!). PpBb x ppBb 29 pouched kangaroos with bushy tails 32 unpouched kangaroos with bushy tails 11 pouched kangaroos with skinny tails 8 unpouched kangaroos with skinny tails For a chi-squared analysis, how many unpouched kangaroos with skinny tails do you expect? Type your answer...In watermelons, bitter fruit (B) is dominant over sweet fruit (b), andyellow spots (S) are dominant over no spots (s). The genes for these two characteristics assort independently. A homozygous plant that has bitter fruit and yellow spots is crossed with a homozygous plant that has sweet fruit and no spots. The F1 are intercrossed to produce the F2. Q. What are the phenotypic ratios in the F2?In mice black fur (B) is dominant over white fur (b). What is the phenotypic percentage of offspring from a cross between a heterozygous black mouse and a white mouse? 100% black; 0% white 50% black; 50% white 100% gray 75% black; 25% white
- In watermelons, bitter fruit (B) is dominant over sweet fruit (b), and yellow spots (S) are dominant over no spots (s). The genes for these two characteristics assort independently. A homozygous plant that has bitter fruit and yellow spots is crossed with a homozygous plant that has sweet fruit and no spots. The F1 are intercrossed to produce the F2. a. What are the phenotypic ratios in the F2? b. If an F1 plant is backcrossed with the bitter, yellow-spotted parent, what phenotypes and proportions are expected in the offspring? c. If an F1 plant is backcrossed with the sweet, unspotted parent, what phenotypes and proportions are expected in the offspring?Feather colour in parakeets is controlled by two genes. Blue colour (B) is dominant over absence of colour (b). Yellow colour (Y) is dominant over absence of colour (y). When a B allele and a Y allele are present, a green parakeet is produced. What is the percent probability of obtaining a yellow or blue parakeet when a heterozygous green parakeet is crossed with a white parakeet? Show all work including Punnett square and explanation. Record your answer as a whole number rounded to one decimal place.In horses, black color (B) dominates chestnut color (b). Trotting gait dominates pacing gait (t). A cross is made between a horse homozygous for both black color and pacing gait and a horse homozygous for both chestnut color and trotting gait. The probable genotype of the offspring resulting from such a cross is and the probable phenotype of the offspring resulting from such a cross is .
- In peaches fuzzy skin (F) is completely dominant to smooth, nectarine skin (f). The gland at the base of the leaves is controlled by a separate, unlinked locus. The gland in heterozygotes is round, while homozygotes with the Go allele have oval glands, and homozygotes with the the GN allele have no glands. A homozygous fuzzy, no-gland peach variety is bred to a smooth, oval-gland variety. The F1 offspring are interbred to produce the F2 generation. What fraction of the F2 offspring will be smooth with round glands? A) 3/16 B) 7/16 C) 4/16 D) 6/16 E) 9/16 F) 2/16 G) 12/16 H) 1/16In watermelons, bitter fruit (B) is dominant over sweet fruit (b), and yellow spots (S) are dominant over no spots (s). The genes for these two characteristics assort independently. A homozygous plant that has bitter fruit and yellow spots is crossed with a homozygous plant that has sweet fruit and no spots. The F1 are intercrossed to produce the F2. What will be the phenotypic ratios in the F2? 2. If an F1 plant is backcrossed with the bitter, yellow-spotted parent, what phenotypes and proportions are expected in the offspring? 3. If an F1 plant is backcrossed with the sweet, non-spotted parent, what phenotypes and proportions are expected in the offspring? 4. In cats, curled ears (Cu) result from an allele that is dominant over an allele for normal ears (cu). Black color results from an independently assorting allele (G) that is dominant over an allele for gray (g). A gray cat homozygous for curled ears is mated with a homozygous black cat with normal ears. All the F1…You cross a black dog which is homozygous dominant for two genes B and E (B determines pigment production and E determines deposition of pigment) with a yellow dog homozygous recessive for both genes and then intercross the F1, you obtain a progeny in the ratio of 9 (black) : 3 (brown) :4 (yellow). Explain the results (specially why you don't get the usual 9:3:3:1 ratio)