Hey guys,
I'm a bit confused as to whether or not this would be an example of epistasis and how to trace the pedigree to find out which it the dominant trait. My TA hasn't posted a key to this practice question so I was hoping someone could explain it to me!
Thanks,
Trent
When true-breeding long-haired, black-eyed guinea pigs are mated with shorthaired, pink-eyed guinea pigs the progeny all have medium length hair and black-eyes. When the F1 progeny are mated back to the shorthaired, pink-eyed parents, what progeny would be expected?
Hi Trent,
It seems like you're working on a genetics problem. Let's break it down and discuss epistasis and tracing the pedigree.
Epistasis refers to a genetic phenomenon in which the effect of one gene masks or modifies the effect of another gene. In this case, we have guinea pigs with different coat lengths (long-haired vs. short-haired), eye colors (black-eyed vs. pink-eyed), and resulting progeny with medium hair length and black eyes.
To determine if this is an example of epistasis, we need to examine how the genes for hair length and eye color interact. If the genes for hair length and eye color are independent and simply segregate according to Mendelian genetics, then this would not be an example of epistasis. However, if the genes interact in a way that influences the expression of one another, then it could be considered epistasis.
Tracing the pedigree can help us understand the inheritance patterns in this case. Since we're looking at the F1 generation mating with one of the parent types (shorthaired, pink-eyed guinea pigs), we can use Punnett squares to determine the possible genotypes and phenotypes of the progeny.
Here's an example Punnett square for the F1 x shorthaired, pink-eyed mating:
| P | p |
---|---|---|
P | PP | Pp |
p | Pp | pp |
In this case, "P" represents the dominant allele for hair length (long-haired), and "p" represents the recessive allele for hair length (short-haired). We know that all the F1 progeny have medium hair length, so their genotype must be "Pp."
The same Punnett square can be used to determine the genotypes and phenotypes of the progeny produced by crossing the F1 generation with the shorthaired, pink-eyed parents.
If we assume that the genes for hair length and eye color segregate independently, we can use another Punnett square to combine the genotype of the F1 progeny with the genotype of the shorthaired, pink-eyed parents:
| Pp | pp |
---|---|---|
Pp | PP | Pp |
p | Pp | pp |
From this Punnett square, we can see that the expected progeny would have a 50% chance of being long-haired (PP), 50% chance of being medium-haired (Pp), and all would have black eyes. The presence of the recessive allele for hair length (p) would result in both long and medium hair lengths.
To summarize, based on the information provided, the expected progeny from mating the F1 generation with the shorthaired, pink-eyed parents would have a mix of long-haired and medium-haired guinea pigs, all with black eyes.
I hope this helps you understand the problem better! Let me know if you have any further questions.