M
mattinuk3
Guest
Hey guys I'm new to the forum. Just wanna say that I know a lot about Mendellian genetics because I'm a little more than half way through a Biology B.S degree. I can do my best to help explain if you guys have some questions. The sad part is that a huge chunk of that information up there can be explained in 2 minutes by somebody doing a Punnet square in front of you while explaining the dominant/recessive traits, but would take so much time to actually digest it from reading it. All of those odd ratios you see are the results of looking for more than one trait at a time, which gets hairy pretty quick because the plants needed grows exponentially. If you learn to do your own punnet square out to the point where you can do a dihybrid cross (where you look for two traits at once in a given cross) you will be able to to know almost all you need for successful breeding.
BASICALLY, Gregor Mendel is the man. Through his experiments he found that (Law of Independent Segregation) each parent has 2 copies of a gene, and only passes on one in its gamete or in our case pollen/ovule. The gene passed is random, but since they make so many gametes, it comes out to 50%/50% of each gene.
This is where dominant/recessive traits come in to play. Usually for a trait, it uses the same letter, but one capitalized and one lower case. Lets say bubblegum scent is recessive to pine scent, therefore b is bubblegum and B is pine. Since each parent has 2 copies of a gene, it can have either BB, Bb, or bb. Following so far?
Now, I insert handy youtube video and let you see how to do a punnett square. This guy looks like he might use a punnet square for the same thing we do lol. http://www.youtube.com/watch?v=r0CtS1Ib1Kw
Once you do the monohybrid punnet square, you can see that even if only one parent has a dominant trait, all of the f1 show it. That is why, unless plants are related somewhere in their pedigree, f1 plants always grow uniform with one phenotype. If we crossed a hypothetical BB (ex Mexican) with a bb(ex. Afghanistan), every single f1 would be heterozygous, or Bb. Every single one of these plants carry the recessive traits, but since they also carry the dominant traits the recessive traits are masked.
Now the f2 is where it gets fun because this is when the recessive traits start to show up. If we inbreed the Bb f1s in the example from above, the offspring will have a 3:1 ratio of dominant:recessive. (do this one with the punnett square!) This one is important and is where Mendel found his famous 3:1 ratio. After doing the punnett square, one will be BB, 2 will be Bb, and one will be bb. At the f2 stage you find out which traits are recessive because traits you didn't see in f1 will show up, and be in a much smaller proportion compared to the dominant (f1 phenotype) plants. These recessive traits may be beneficial or they may be harmful because they are just a random shuffling of the genetic dice.
If you want to check two traits at once, you need to do a dihybrid cross. This is a 4x4 punnett square. The odds of getting a plant with two recessive traits is 1:16. If you are looking for 3 recessive traits the odds are 1:64. Assuming your 1:64 plant cracks its seed, survives, and has 3 beneficial recessive traits, you still have to find the other sex with all 3 traits too. This is why I said, for the purposes of what we're doing, a dihybrid cross is all we need to work with. And, even then, maintaining a successful breeding program while looking for two recessive traits at once would probably require growing 100 plants at a time and soooo much time. This is what I would eventually like to do some day though.
Anyways, I don't know if I'm just preaching to the choir, but soak that in and try to do a few punnet squares with some traits you have in mind from some plants you've grown recently. Then, hit me with some questions if you have them. I wanna break down the science gibberish to better the marijuana gene pool!
BASICALLY, Gregor Mendel is the man. Through his experiments he found that (Law of Independent Segregation) each parent has 2 copies of a gene, and only passes on one in its gamete or in our case pollen/ovule. The gene passed is random, but since they make so many gametes, it comes out to 50%/50% of each gene.
This is where dominant/recessive traits come in to play. Usually for a trait, it uses the same letter, but one capitalized and one lower case. Lets say bubblegum scent is recessive to pine scent, therefore b is bubblegum and B is pine. Since each parent has 2 copies of a gene, it can have either BB, Bb, or bb. Following so far?
Now, I insert handy youtube video and let you see how to do a punnett square. This guy looks like he might use a punnet square for the same thing we do lol. http://www.youtube.com/watch?v=r0CtS1Ib1Kw
Once you do the monohybrid punnet square, you can see that even if only one parent has a dominant trait, all of the f1 show it. That is why, unless plants are related somewhere in their pedigree, f1 plants always grow uniform with one phenotype. If we crossed a hypothetical BB (ex Mexican) with a bb(ex. Afghanistan), every single f1 would be heterozygous, or Bb. Every single one of these plants carry the recessive traits, but since they also carry the dominant traits the recessive traits are masked.
Now the f2 is where it gets fun because this is when the recessive traits start to show up. If we inbreed the Bb f1s in the example from above, the offspring will have a 3:1 ratio of dominant:recessive. (do this one with the punnett square!) This one is important and is where Mendel found his famous 3:1 ratio. After doing the punnett square, one will be BB, 2 will be Bb, and one will be bb. At the f2 stage you find out which traits are recessive because traits you didn't see in f1 will show up, and be in a much smaller proportion compared to the dominant (f1 phenotype) plants. These recessive traits may be beneficial or they may be harmful because they are just a random shuffling of the genetic dice.
If you want to check two traits at once, you need to do a dihybrid cross. This is a 4x4 punnett square. The odds of getting a plant with two recessive traits is 1:16. If you are looking for 3 recessive traits the odds are 1:64. Assuming your 1:64 plant cracks its seed, survives, and has 3 beneficial recessive traits, you still have to find the other sex with all 3 traits too. This is why I said, for the purposes of what we're doing, a dihybrid cross is all we need to work with. And, even then, maintaining a successful breeding program while looking for two recessive traits at once would probably require growing 100 plants at a time and soooo much time. This is what I would eventually like to do some day though.
Anyways, I don't know if I'm just preaching to the choir, but soak that in and try to do a few punnet squares with some traits you have in mind from some plants you've grown recently. Then, hit me with some questions if you have them. I wanna break down the science gibberish to better the marijuana gene pool!
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