ilovehaein

                                                                                                                                                           Report-05 2015-11-2

This time, we are here to talk about to see if the 3:1 ratio still works when there are two different characters which have a phenotype when shown.
And I'll tell you one thing, you'll see something familiar when you know it.
Since i'm assuming that you've seen the last writing and had a lot of thoughts to it, I only have one thing to say - there are two characters, they are the texture and color of the seed. The round gene is R, wrinkled is r, yellow is Y, and green is y.
NOTE: R and Y is dominant and r and y is recessive.
If we are all set, let's go for it!

Let's say that we have two pure(original) pea plants that is round and yellow, and are wrinkled and green.
Since we said that the pea plants are "pure", that means that there are no heterozygotes in the genotypes. Am I right?
Now, here is what we are going to have something called "Dihybrid cross".
This word means that we are going to have a hybridization within the alleles that has an complete dominant alleles and complete recessive alleles.
Now with all that information, I'm sure you all might notice what the whole story goes - one pea pod is going to have a gene of RRYY and a another pea pod is rryy.

Right now, I know you now got frustrated with this whole "two genes in one" kinda stuff, but this is just a fancy way of saying "I have a RR gene and a YY gene inside my individual chromosomes in my cell" and vice versa.
For a picture(If you want,) it's like this -
RRYY { (RR) (YY) } rryy { (rr) (yy) }
The {} is a cell, and () is a gene locus(~location).

Now, let's see the first picture of Mendel's second law. When you have a RRYY and a rryy, the first thing to do is to know what kind of gene possibilities you can have when you calculate the genotypes of the F1 generation with the punnet square.
Since we know that neither the characters 'texture' and 'color' cannot be apart from the pea plant's gene, the RRYY is only going to have a RY gene passed out to the offspring while the rryy is going to pass down the ry gene only.
This means by definition that we are not going to have any biological errors, the F1 generation is going to have a genotype of RrYy and a phenotype of round and yellow pea plants, because R and Y are dominant to the r and y.
Then, what about the F2 generation? What possible phenotypes could they produce to us?

We do know that the pea plants are going to self-fertilize, which means that the F1 generation pea plants are going to fertilize with each other. This means a RrYy gened pea plant and the exact gened RrYy pea plant is going to fertilize for a F2 generation.
Let's look at the punnet square at the bottom of the picture.
This time since we are dealing with a genotype of RrYy, we are going to have 4 outcomes - RY, Ry, rY, ry.
if we punnet square them, we would get nine genotypes of the F2 generations - RRYY, RRYy, RRyy, RrYY, RrYy, Rryy, rrYY, rrYy, rryy.
But unlike the genotypes, we would get four phenotypes of the F2 generations because the genotypes that has at least one to two dominant allele in a character would show the dominant stuff and covers the recessive ones - Round and yellow, wrinkled and yellow, round and green, wrinkled and green.
*NOTE*: When you look at the punnet square, you would see that there are 16 genotypes, not 9, this is just because there are a bit more of the same genes(example: there are two pea plants that has the genotype RRYy).

Say, can I give you a question?
What do you think what would happen if the R gene will ALWAYS stick to the Y gene and vice versa? What would happen if this pea plant has self fertilized?
This is sudden, but a noting word. What I meant about the R gene always sticking with the Y gene is that the locus(location) of the R gene and the locus of the Y gene is on the same chromosome.
What's wrong with that kind of style is that when the cells undergo meiosis, the "R and Y" pair and the "r and y" pair will never come apart - therefore it never makes a random combination of the genes, not making any of the offspring a heterozygote. That kind of genotyped pea plant will always have either a RY or a ry to give to the offspring when self fertilized, and we cannot say this as Mendel's law of independent assortment AKA "Mendel's second law".
Go to this site for the same information (Sorry that it's in korean, english users!) http://study.zum.com/book/18094

So now that we have proved that we cannot have the texture charactered gene and the color charactrized gene in the same chromosome(or "sticking together forever like best pals."), I would give you a easy question for you -
Do you think the answer will be the same if we had more Ry genes to be put in the punnet square? the answer is NO! we cannot have more or lesser genes then the average ones we already have.
Therefore, we can say that the ratio of the RY, Ry, rY, ry gene possibilities 'mix' is 1 : 1 : 1 : 1
Look at the second picture, if you can see the purple circles with down arrows and a number 1 on top of them, that is the stuff I was talking about the whole time.
This will help you too.https://www.ndsu.edu/pu…/~mcclean/plsc431/mendel/mendel3.htm

Now, look at the third picture that has a lot more red text then any of the other pictures. Because we will see some obvious but reasonable stuffs.
Let's remember the thoughts, The parent pea plants are RRYY and rryy, which can only hand down RY and ry to the F1 generation.
The F1 generation is RrYy, so it can give out either the RY, Ry, rY, ry genes to it's second generation(F2). Look at the purple circle with the RY, Ry, ry, ry genes, do you see something familiar?
That's right! The genes RY and ry is handed down from the parent genes RRYY and rryy. So we can call that the RY and ry gene are parental genes.
Then, what about the Ry and rY genes?
Those genes are from the F1 generation which is a union of the two parental genes RRYY and rryy.

Therefore, we could say that the Ry and rY genes are recombinant genes.
Then, what is the ratio between the parental and recombinant genes?
Let's think about the information we currently have for this question.
We do know that the ratio between RY, Ry, rY, ry is 1 : 1 : 1 : 1,
And we know that RY and ry are parental genes, and Ry and rY are recombinant genes.
This information we currently have can tell us that the ratio between the parental genes : recombinant genes is going to be the same as the ratio between RY and ry : Ry and rY.
This gives us a answer of 2 : 2 because all of the ratios of the genotypes for the punnet square is 1.
if we simplify this, it will be 1 : 1, or 50% over 50%. That is the ratio.

We did the ratio of the purple circles, now let's do the ratio of the punnet square.
we can see that 9 of the 16 possibilities of the pea plants are round and yellow, and 3/16 is wrinkled and yellow, another 3/16 is round and green, and the last 1/16 is wrinkled and green.
in Mendel's first law, or Mendel's law of dominance and inheritance shows us a ratio of 3 : 1 whereas Mendel's second law (Mendel's law of independent assortment) gives us a ratio of 9 : 3 : 3 : 1.
You might think that 3 : 1 and 9 : 3 : 3 : 1 has nothing in connection, but look at the punnet square. Because i'll give you two questions.
What is the ratio between the round and wrinkled pea plants? It's 3 : 1.
What is the ratio between the yellow and green pea plants? It's also a 3 : 1 ratio.
So, the 3 : 1 ratio still works, but only the difference it has is that since we have two traits(characters), we have to kinda "Double" the 3 : 1 ratio into a 9 : 3 : 3 : 1 ratio.
Check out for wikipedia, the free encyclopedia about Mendel's laws for more information!
https://en.wikipedia.org/wiki/Mendelian_inheritance

Well, this is all for today's dihybrid cross of two pea plants that each has two character traits - texture and color.
I'm just still 12, so advanced biology users! Please tell me if there is anything wrong with my article, thank you!

By Jane Kim
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