1.) In a plant having two carried alleles for the color of a flower in a gene, with P for purple and p for white, the three possible combinations which might exist in any one plant are PP making a purple plant, pp to make a white plant, or Pp resulting in a ‘hybrid’ plant.
2.) Out of the genotypes PP, Pp, pp, the resultant flower colors are (as described above in exercise 1) are purple (for PP,) purple or purplish-white (for Pp- likely purple as it is dominant, or a mixture of the colors,) or white (for the case of pp.) PP and pp, the purple and white flowers, are referred to as homozygous. In the case of PP this is homozygous dominant, and in the case of pp this is homozygous recessive. The case of Pp must be considered different, and is classified and heterozygous.
3.) For a plant that is Pp, two gametes are produced: P and p. For a plant that is PpTt, four gametes are produced: PT, pT, Pt, and pt. For a plant that is PpTtYy, eight gametes are ultimately produced: PTY, PTy, Pty, pTy, ptY, PtY, pTY, and pty.
4.) To address the first question we will consider the Punnett square to cross the heterozygous purple plant and a white flowered plant. See the Punnett square drawn here and labeled “Figure 1” below.
| |p |p |
|P |Pp |Pp |
|p |pp |pp |
Figure 1: Punnett square crossing heterozygous purple plant and a white flowered plant
To address the second question, we will consider the genotypes produced from this mating. There are shown in the Punnett square, and overall there are two total genotypes (as there are two cases of the Pp genotype so technically there are not four in total.) These genotypes are Pp and pp, similar to the first question however here in relation to the Punnett square breakdown and without a dominant PP. To address the third question, we will consider the phenotype, or the color in this case. The genotype pp (homozygous recessive) will possess the phenotype of white. Meanwhile, the genotype Pp (heterozygous) may possess the phenotype of purple and white in mixture, while purple is the dominant trait in the gamete P. To address the fourth question, we consider the homozygous versus heterozygous traits. As mentioned, pp is homozygous (homozygous recessive,) while the Punnett square reveals two instances of the heterozygous Pp. Thus, fifty percent of the offspring would be classified as being homozygous (recessive.)
5.) To address the first question, we will calculate the Punnett square while crossing two pink flowers, R1R2. The results are shown below in “Figure 2.”
| |R1 |R2 |
|R1 |R1R1 |R1R2 |
|R2 |R1R2 |R2R2 |
Figure 2: Punnett square calculation for pink flowers.
The results from this Punnett square will be shown when considering the other questions in this problem. Now we will address the second question. While R1R1 is red, R2R2 is white, and R1R2 is pink, we can see the result of crossing two pink flowers has resulted in 25 percent red flowers, 25 percent white flowers, and 50 percent pink flowers in offspring. To address the third question, we will use a Punnett square to cross a roan shorthorn bull (denoted R’R) with a white cow (denoted R’R’) and consider the results in “Figure 3.”
| |R’ |R | |R’ |R’R’ |R’R | |R’ |R’R’ |R’R |
Figure 3: Punnett square calculation for crossing a roan shorthand bull (R’R) with a white cow (R’R’).
Considering the above phenotypes (and being mindful that the RR type comes from a red coat bull,) we can see the percentages of roan shorthorn bulls with roan coats, white coats, and red coats. To address the fourth question, the Punnett...
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