Hi there!
I have this question for my bio/genetics homework, and I can't seem to figure it out.
It reads, "A true-breeding red snapdragon is crossed to a true-breeding white snapdragon. The F1 progeny are all red. When the F1 is selfed, the following F2 progeny are observed:
1850 red (55%)
670 pink (20%)
831 white (25%)
(3351 total)
Give the appropriate genotypes of the parents, F1 and F2 progeny and name the type of genetic interaction.
Additionally, if random mating were to happen between the gametes of the F2 generation, what phenotype classes would you expect in the resulting flowers?"
At first I thought it was incomplete dominance, but that wouldn't make sense of true Red was RR and true white was rr. Is there something I'm missing?
This is an example of recessive epistasis. The ratios are 9:3:4. Red flowers genotype would be R/-;W/-. Pink flowers would be r/r;W/- and white flowers would have either R/-;w/w or r/r;w/w. The f1 progeny being all red resembles one of the parents, which is a clue for epistasis. If it was incomplete dominance then a cross between RR and ww would result in all pink F1 generation. Hope this helps!
To determine the genotypes and genetic interaction in this scenario, we need to use the principles of Mendelian genetics. Let's break down the information step by step.
1. A true-breeding red snapdragon is crossed to a true-breeding white snapdragon.
- True-breeding means that the parents are homozygous for a specific trait. In this case, red snapdragon (RR) is crossed with white snapdragon (rr).
2. The F1 progeny are all red.
- When the true-breeding red parent (RR) is crossed with the true-breeding white parent (rr), all the offspring in the F1 generation are red. This suggests that there is dominance involved in the inheritance pattern.
3. When the F1 is selfed, the following F2 progeny are observed:
- 1850 red (55%)
- 670 pink (20%)
- 831 white (25%)
- (3351 total)
To analyze the F2 progeny, we need to consider the possible genotypes of the F1 generation. Since the F1 progeny are all red (dominant phenotype), it indicates that they must be heterozygous.
Let's assign a symbol for the red color allele (R) and the white color allele (r) to represent the genotype:
- RR: Red snapdragon (true-breeding red)
- rr: White snapdragon (true-breeding white)
- Rr: F1 progeny (red, as they are heterozygous)
Now, let's determine the possible genotypes and phenotypes of the F2 progeny:
- RR (25%): These are the progeny with the genotype RR, which are red in color due to dominance.
- Rr (50%): These are the progeny with the genotype Rr, which are also red in color because the red allele is dominant over white.
- rr (25%): These are the progeny with the genotype rr, which are white in color.
From the given data, we can see that the observed percentages match the expected ratios for a Mendelian inheritance pattern with complete dominance.
Therefore, the appropriate genotypes are as follows:
- True-breeding red parent: RR
- True-breeding white parent: rr
- F1 progeny: Rr
- F2 progeny: RR, Rr, rr
The type of genetic interaction involved in this case is complete dominance, with the red allele (R) being dominant over the white allele (r).
Now, if random mating were to occur between the gametes of the F2 generation, we can use the principle of Mendelian genetics to predict the possible phenotype classes in the resulting flowers. As explained earlier, the F2 generation has three possible genotypes: RR, Rr, and rr.
When random mating occurs, the possible combinations of these genotypes are as follows:
- RR x RR: 25% RR (red)
- RR x Rr: 50% RR (red) and 50% Rr (red)
- RR x rr: 100% Rr (red)
- Rr x Rr: 25% RR (red), 50% Rr (red), and 25% rr (white)
- Rr x rr: 50% Rr (red) and 50% rr (white)
- rr x rr: 100% rr (white)
Therefore, the resulting phenotype classes in the flowers would be red and white, with various ratios depending on the genotype combinations resulting from random mating between the gametes.
I hope this explanation helps you understand the problem and its solution! Let me know if you have any further questions.