DRAW IT Two pea plants heterozygous forthe characters of pod color and pod shape arecrossed. Draw a Punnett square to determine thephenotypic ratios of the offspring

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96% of Florida panther population has normal tails.  4% of the Florida panther population has the crook in the tail. Crooked tails are recessive.      Find the dominant allele frequency and the recessive allele frequency.  Construct a Punnett Square that would illustrate the phenotypic outcome of the a parent who is homozygous for a normal tail and homozygous for a crooked tail.  Include a Punnett square illustrating the cross.  Include a “key” or “legend” describing the pattern of inheritance for the trait, the genotypes of the parents, and the phenotypic percentages of the outcome of the cross.

Tenera is grown for consumption seeds not as a source of oil palm seeds so it is not recommended to plant uncertified fruit/seeds, why? Explain by making crosses starting from parental (Dura and Pisifera) to F2 complete with pictures of the fruit and its gene structure and the cross chessboard.

draw a punnett square for both parents are dominate tall, name the 4 possible offspring.

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Incomplete dominanceIn Mendel's experiments, if the gene for tall (T) plants was incompletely dominant overthe gene for short (t) plants, what would be the result of crossing two Tt plants? Explain result.

Most of the feathers of erminette fowl are light colored, with an occasional black one, giving a fleckedappearance. A cross of two erminettes produced a totalof 48 progeny, consisting of 22 erminettes, 14 blacks,and 12 pure whites. What genetic basis of the erminette pattern is suggested? How would you test yourhypotheses?

non OBYTOT SEI deutet bos nuol 3. Based on the total number of kernels of corn you counted, calculate the expected number for each phenotype if the parents of this corn were both RrTt. Show how you calculate these numbers below and en transfer them to the observation table (155 Kernels) then INO 4 4. Using chi-square analysis to compare the actual phenotype and expected phenotype numbers determine if you can statistically accept that both parents have the genotype RrTt. Your chi square data will be recorded to helow. nb number) which allows for