Assume that two pigments, red and blue, mix to give the normal purple color of petunia petal. Separate biochemical pathways synthesize the two pigments, as shown in the top two rows of the accompanying diagram. "White" refers to compounds that are not pigments. (Total lack of pigment results in a white petal.) Red pigment forms from a yellow intermediate that is normally at a concentration too low to color petals.
Pathway I ... → white1 →E blue
Pathway II ... → white2→A yellow→B red
C↑
Pathway III ... →white3 →D white4
( the letters should be on top of the arrows, and the numbers are subscripts)
A third pathway, whose compounds do not contribute pigment to petals, normally does not affect the blue and red pathways, but, if one of its intermediates(white) should build up in concentration, it can be converted into the yellow intermediate of the red pathway. In the diagram, the letters A through E represent enzymes; their corresponding genes, all of which are unlinked, may be symbolized by the same letters. Assume that wild-type alleles are dominant and encode enzyme function and that recessive alleles result in a lack of enzyme function. Deduce which combinations of true-breeding parental genotypes could be crossed to produce F2 progeny in the following ratios:
a. 9 purple: 3 green: 4 blue b. 9 purple: 3 red: 3 blue: 1 white c. 13 purple: 3 blue d. 9 purple: 3 red: 3 green: 1 yellow
To determine the combinations of true-breeding parental genotypes that could produce the ratios given, we need to analyze the pathways and the interactions between genes.
First, let's assign the letters to represent the alleles of each gene:
- Gene A: allele A (wild-type) and allele a (recessive)
- Gene B: allele B (wild-type) and allele b (recessive)
- Gene C: allele C (wild-type) and allele c (recessive)
- Gene D: allele D (wild-type) and allele d (recessive)
- Gene E: allele E (wild-type) and allele e (recessive)
Now let's evaluate each case:
a. 9 purple: 3 green: 4 blue
For purple petals, both the blue and red pathways need to be functioning. Therefore, the parental genotypes must have at least one functional allele for gene E.
Possible parental genotypes that can produce this ratio:
- AA BB CC DD EE (wild-type for all genes)
- Aa Bb CC DD EE or Aa BB Cc DD EE (carriers of recessive alleles for genes C or B, respectively)
b. 9 purple: 3 red: 3 blue: 1 white
For purple petals, both the blue and red pathways need to be functioning. For red petals, only the red pathway needs to be functioning. The white petal phenotype suggests that there may be a lack of pigment production, possibly due to a recessive allele at one of the genes involved in pigment synthesis.
Possible parental genotypes that can produce this ratio:
- AA BB CC Dd ee (recessive allele d in gene D is required for white petals)
- Aa Bb CC Dd EE or Aa BB Cc Dd EE (carriers of recessive alleles for genes C or B, or both, respectively, and recessive allele d in gene D)
c. 13 purple: 3 blue
For purple petals, both the blue and red pathways need to be functioning. However, the absence of red petals suggests that there may be a recessive allele at gene B, preventing the conversion of yellow pigment to red pigment.
Possible parental genotypes that can produce this ratio:
- AA bb CC DD EE (recessive allele b in gene B is required for blue petals)
- Aa bb CC DD EE (carrier of recessive allele b in gene B)
d. 9 purple: 3 red: 3 green: 1 yellow
For purple petals, both the blue and red pathways need to be functioning. For red petals, only the red pathway needs to be functioning. Green petals indicate a lack of blue pigment production, likely due to a recessive allele at gene E.
Possible parental genotypes that can produce this ratio:
- AA BB CC Dd ee (recessive allele d in gene D is required for green petals)
- Aa BB CC Dd ee (carrier of recessive allele d in gene D)
- AA bb CC DD EE (recessive allele b in gene B is required for yellow pigment production)
- Aa bb CC DD EE (carrier of recessive allele b in gene B)
Please note that these are possible combinations based on the given information, but there may be other combinations that could also produce the same ratios.
To deduce which combinations of true-breeding parental genotypes could produce the given ratios of progeny, we need to analyze the pathways and understand how the pigments are synthesized.
Let's break down the information given in the question:
1. Pathway I: It starts with a compound that we'll call "white1" and leads to the blue pigment.
2. Pathway II: It starts with a compound that we'll call "white2," which is converted into a yellow intermediate (A) and then further into the red pigment (B).
3. Pathway III: It starts with a compound that we'll call "white3" and leads to a compound that we'll call "white4."
Now, let's analyze each scenario:
a. 9 purple: 3 green: 4 blue
This ratio suggests that there are three different phenotypes: purple, green, and blue. Since blue is the result of the Pathway I, we can deduce that one of the parental genotypes must have a genotype for blue pigment synthesis. The purple color is produced by the combination of red and blue pigments, so another parental genotype must carry at least one gene for red pigment synthesis.
Therefore, the parental genotypes could be:
- Parent 1: Blue (Ee) - genotype for pigment synthesis in Pathway I
- Parent 2: Red (Bb) - genotype for pigment synthesis in Pathway II
- F1 progeny: EeBb (heterozygous for both pigments)
When the F1 progeny cross, they may assort independently into different genotypes, leading to the expected ratio of 9 purple: 3 green: 4 blue in the F2 generation.
b. 9 purple: 3 red: 3 blue: 1 white
This ratio suggests that there are four different phenotypes: purple, red, blue, and white. Since white refers to compounds that are not pigments, it is likely that one of the parental genotypes carries recessive alleles for both pigment pathways.
Therefore, the parental genotypes could be:
- Parent 1: Blue (Ee) - genotype for pigment synthesis in Pathway I
- Parent 2: White (ddee) - lacks enzyme function for both pigments
- F1 progeny: EeDd (heterozygous for blue pigment but lacking red pigment)
When the F1 progeny cross, they may assort independently into different genotypes, resulting in the expected ratio of 9 purple: 3 red: 3 blue: 1 white in the F2 generation.
c. 13 purple: 3 blue
In this scenario, there are two phenotypes: purple and blue. The absence of any white or green phenotypes suggests that both parental genotypes must carry at least one functional gene for pigment synthesis.
Therefore, the parental genotypes could be:
- Parent 1: Blue (Ee) - genotype for pigment synthesis in Pathway I
- Parent 2: Red (Bb) - genotype for pigment synthesis in Pathway II
- F1 progeny: EeBb (heterozygous for both pigments)
When the F1 progeny cross, they may assort independently into different genotypes, leading to the expected ratio of 13 purple: 3 blue in the F2 generation.
d. 9 purple: 3 red: 3 green: 1 yellow
This ratio suggests that there are four different phenotypes: purple, red, green, and yellow. Since yellow is an intermediate in the Pathway II and is not directly related to green pigment synthesis, it is likely that one of the parental genotypes carries a recessive allele for yellow synthesis.
Therefore, the parental genotypes could be:
- Parent 1: Red (Bb) - genotype for pigment synthesis in Pathway II
- Parent 2: Yellow (Aa) - genotype for yellow pigment synthesis
- F1 progeny: AaBb (heterozygous for both pigments)
When the F1 progeny cross, they may assort independently into different genotypes, leading to the expected ratio of 9 purple: 3 red: 3 green: 1 yellow in the F2 generation.
Remember, these are only proposed genotypes based on the given ratios. To confirm these predictions, it would be necessary to perform the actual crosses and observe the offspring.