Last time, we learned about enzymes and their functions.

Now, we are here to talk about how to use some mathematics to understand more about their interactions! Shall we start?

Enzyme kinetics is the study of chemical reactions catalyzed by enzymes.

It concentrates a lot on the speed - Or I should say "Rate".

By doing this, we can study how we can alter the rate of reaction or it's functionality.

In enzyme kinetics, we have some terms to learn about before we do anything else. The first one is this: E + S ↔ ES → E + P

Where E responds to the enzyme, S as the substrate, ES as the ES complex, and P as the product. The enzyme - substrate complex can rewind back to an enzyme and its substrate. The Enzyme and it's product can do the same too.

However, because products are usually very stable, they hardly rewind back.

Assuming that you know kinetics, let's call the forward reaction of E + S → ES "Reaction 1". And "Reaction 2" for the forward reaction of ES → E + P.

Then, the rate of our reactions will be like this:

Rate1 = K1[E][S] and Rate2 = K2[ES]

Where Rate1 is the rate of reaction 1 and vice versa.

Same for K1 and K2. Got it? Ok.

Next is something called "Vmax". And... What is it?

Say that we have an enzyme that can catalyze 10 reactions per second.

If we had four of them, what would be the maximum rate?

40 reactions per second, right? That's what we call "Vmax". 

The max rate that the enzyme can catalyze per second. That's Vmax.

By the way, Vmax will always occur in every situation. Keep that in mind please.

Next is a diagram where you plot the reaction rate(V) and the substrate concentration. 

We start with no reaction rate and no substrate.

Then as we add more substrate, our rate goes up, and it becomes more flatter in the end, approaching a certain value. The point where it flattens out is when the enzyme is saturated with substrate.

Which means it's reached it's Vmax.

Now, if you look at the y - axis, you can see 1/2 of the Vmax. 

It's corresponding X axis is also shown as well too.

This X axis - Or substrate concentration is called Km.

Km, is also called the Michaelis Constant. We will talk about why it's called like that later. So this Km, is the substrate concentration when the rate is 1/2 Vmax. When people talk about this, some say that Km is 1/2 Vmax.

That is Wrong! I mean like, Really WRONG!

Km is a Substrate Concentration. Remember this and you'll be fine! 

Next is the Michaelis Menten Equation!

E + S ↔ ES → E + P

Since we have reaction 1 as the forward for E + S ↔ ES, let's get ourselves a reverse reaction and call it "Reaction -1". We can make a "Reaction -2" for the same reason, but as I said - "Products hardly reverse back". Let's sign that out.

Now, we are going to have something called the Steady - State Assumption.

At steady state, [ES] is constant. Meaning that the rate of formation of [ES] is equal to the rate of dissociation of [ES].

K1[E][S] = K-1[ES] + K2[ES]

Let's group the right side with [ES].

K1[E][S] = (K-1+ K2)[ES]

Now let's move the rate constants on the right and the others on the left!

We will get [E][S]/[ES] = (K-1 + K2)/K1

Turns out, is that this Km is actually (K-1 + K2)/K1!

Then we will get Km =[E][S]/[ES]

Let's move the [ES] from the right to the left side.

[ES]Km = [E][S]

Now let's define a new term called [E]T or "total enzyme" as [E]T = [E] + [ES]

Then we can substitute the [E] from the right to [E]T - [ES]. 

That gives us [ES]Km = ([E]T - [ES])[S].

If we distribute the right side, it's: [ES]Km = [E]T[S] - [ES][S].

By moving the [ES][S] to the left side and grouping the [ES] together, we get

[ES](Km + [S]) = [E]T[S].

If we leave [ES] on the left side only, then we get [ES] = [E]T[S]/(Km + [S]).

Our rate is equal to the rate of formation of product, Or K2[ES].

So Vo = K2[ES]. At high [S], The enzyme will be saturated with substrate. So the Vo = Vmax. And since all the substrate are bonded with the enzyme, there are no free enzymes in the solution. Meaning that [E]T = [ES] as well.

So it also means that Vmax = K2[E]T as well.

Let's multiply K2 on both sides of [ES] = [E]T[S]/(Km + [S]).

Then we get K2[ES] = K2[E]T[S]/(Km + [S]).

By using our terms above, we get Vo = Vmax[S]/(Km + [S]).

This, is the Michaelis Menten Equation!

In our usual graph, which is in a hyperbolic curve, is not that quite accurate.

For example, if you got the Vmax wrong because of where the curve was headed to, then you would get the Km wrong and everything!

This is where the Lineweaver - Burk plot comes in.

It's derivation is quite simple. 

First, get our Michaelis Menten equation Vo = Vmax[S]/(Km + [S]).

Second, take the inverse of it, making 1/Vo = (Km + [S])/Vmax[S].

Third, rewrite the equation to 1/Vo = Km/Vmax[S] + [S]/Vmax[S].

Simplifying the equation gives us 1/Vo = Km/Vmax * 1/[S] + 1/Vmax.

And that... Is the same way of saying y = mx + b!

Which gives us a straight line on the graph shown below.

With this, we can get a more accurate value of Km and Vmax!

The X - Intercept is -1/Km, and the Y - Intercept is 1/Vmax.

And there are the other values shown in the picture above.

In enzyme kinetics, we also learn about Inhibitors as well.

Inhibitors are molecules that decreases the enzyme's activity.

The first type of inhibitor is called a Competitive Inhibitor.

Competitive inhibitors - Like the name says, "Competes" with the substrate for the enzyme. And whoever comes first gets the enzyme.

Normally it's for the active site, but it can also be the Allosteric site as well too.

If we write this out, it will be: E + I ↔ EI, E + S ↔ ES ↔ E + P.

At steady state, our Km will still be [E][S]/[ES]. 

However, now we will have a new term called KI. Which is [E][I]/[EI].

Now let's remove some terms that aren't necessary.

Let's divide Km by KI. Then we get Km/KI = [S][EI]/[ES][I].

Now it looks like we have nothing else to simplify... Or is it?

Let's look at our toolbox of equations! 

Since[EI] is also a part of [E]T, [E]T = [E] + [ES] + [EI].

If we substitute [EI], then Km/KI = ([E]T - [E] - [ES]) [S]/[ES][I].

Simplifying it gives us Km/KI = [E]T[S]/[ES][I] - [E][S]/[ES][I] - [ES][S]/[ES][I]. 

Km = [E][S]/[ES], so [E]/[ES] = Km/[S].

And Vmax = K2[E]T, Vo = K2[ES]. Therefore, Vmax/Vo = [E]T/[ES].

Applying it gives us Km/KI = Vmax/Vo *[S]/[I] - Km/[S] * [S]/[I] - [S]/[I].

Grouping the [S]/[I] and moving it to the left side equals...

Km/KI * [I]/[S] = (Vmax/Vo - Km/[S] - 1).

After this, let's put the Vmax/Vo in one side and everything else in the other.

Vmax/Vo = Km/KI *[I]/[S] + Km/[S] + 1.

By grouping, we get Vmax/Vo = Km(1/KI * [I]/[S] + 1/[S]) + 1.

And then Vmax/Vo = Km([I]/KI + 1) 1/[S] + 1.

Dividing both sides by Vmax is: 1/Vo = Km([I]/KI + 1)/Vmax * 1/[S] + 1/Vmax. 

Lastly, we substitute [I]/KI + 1 into a constant called α.

So it becomes 1/Vo = Km*α/Vmax *1/[S] + 1/Vmax.

This is the Lineweaver burk equation of a competitive inhibitor.

If we graph this, we get the fallowing picture below:

The 1/Vmax stays the same. So there is no difference to the Y - Intercept.

For the X - Intercept however, is different.

If we solve what the X intercept is for the Competitive inhibitor, we get -1/Kmα.

And since α is larger then 1, our X intercept is going to be larger then the original X intercept.

In short, Vmax does not change. However, the apparent Km increases.

How does that happen? The Substrate and the Inhibitor are competing for the enzyme. So if we have a LOT of substrate then the inhibitor, we will be able to reach the original Vmax by beating the inhibitor on the way to the enzyme.

So there is no change to the apparent Vmax.

In order to reach the same Vmax, more substrate is needed to reach the same 1/2Vmax. So the enzyme has a low substrate affinity. Therefore the apparent Km increases.

The reason why I used the word "Apparent" is because the Vmax and the Km itself doesn't change. The inhibitors are making it look like the Vmax and Km changed when it's actually decreasing it's activity.

Vmax and Km are Constants. And Constants stay Constant.

Next are Uncompetitive Inhibitors.

These guys don't "Compete" with the substrate.

Instead, they bind to the Allosteric site when the Enzyme is associated with the substrate. Making an ESI complex. 

Writing it out will show this:

E + S ↔ ES ↔ E + P, ES + I ↔ ESI.

I won't do the LBP derivation all over again. Instead, I'll show it to you directly -

1/Vo = Km/Vmax *1/[S] + α`/Vmax.

Whereas α` = 1 + [I]/K`I. What is this α` and K`I?

K`I is the KI(inhibitor constant) when the inhibitor is uncompetitive,

And α` is α with [I]/K`I in it instead of [I]/KI.

They're just here to distribute which inhibitor they are in.

If we graph it, it will look like this:

It's Y intercept is α`/Vmax. And since α` is also larger then 1, The Y intercept is higher then the original one.

For the X intercept, we get -α`/Km. That makes the X intercept more lower then the original one.

So it means that the apparent Vmax and Km both decreases in an uncompetitive inhibitor. How does that happen?

The uncompetitive inhibitor binds to an ES complex only. So no matter how many substrate you add, the inhibitor is going to stick anyways. So the apparent Vmax decreases.

And as the ESI complexes gets formed, ES is decreasing over time.

So the system tries to balance the equilibrium by accepting more substrate.

This leads to a higher affinity - and thus, low Km.

The third are mixed inhibitors.

These guys are basically a "mix" of the two inhibitors above.

So if we write them out, it's like this:

E + S ↔ ES ↔ E + P, E + I ↔ EI, ES + I ↔ ESI.

It's Vmax will be decreased due to the effects of the ESI.

However for the Km, it can increase, decrease, or stay the same as before. Because it's a mixture of the competitive inhibitor's behavior(increase) and the uncompetitive inhibitor's behavior(decrease).

It's lineweaver burk equation is 1/Vo = Km*α/Vmax * 1/[S] + α`/Vmax.

In a graph, here's what it looks like:

The X intercept is -α`/Km*α. Which can be smaller or bigger then the original X intercept. It depends which is bigger - α or α`.

If α is bigger, then the X intercept is bigger then the original.

If α` is bigger though, then the X intercept is smaller then the original.

In this situation, α is bigger then α`.

For the Y intercept however, is always α`/Vmax. So the Y intercept is always bigger then the original.

The last are Noncompetitive inhibitors.

They are a type of mixed inhibitors. However we call it a noncompetitive inhibitor when α = α`!

It's graph looks like this:

The X intercept is the same as the original one since α = α`.

The Y intercept, of course is α`/Vmax. Which is bigger then the original.

And that is it! The end of Enzyme Kinetics!

I might post some more stuff about things like cooperativity.

Until then, stay tooned!

                                  2016. 5/29

                                  <해인이가>

Image credits section:

Khan academy MCAT videos.

Wikipedia: Main page, Enzyme kinetics section.

Wikibooks: Main page.

EDX: Principles of Biochemistry.

So far, you know that a lot of chemical reactions are happening in your body.

But all of these reactions are slow. 

Even a simple reaction will take a lot of time! 

Because we all know that reactions occurs when two molecules "crash" in the correct order and all those other right conditions.

What makes all these reactions go fast in a second? 

The answer is right here in this article everybody. 

They are Enzymes. Our body's well - known catalyst!

Now, an enzyme is a type of catalyst. A catalyst is a substance which makes reactions go faster. In order for enzymes to do it's job, it needs something to react with. That is the Substrate.

The Active site is the part that enzymes bind with the substrates. 

And the others are all called Allosteric sites. We will talk about them later.

Here's a good thing to know - Most enzymes are proteins. So they are made out of amino acids. However, some proteins are made out of RNA molecules which are called Ribozymes. So not every enzymes are not proteins.

Say, can it be possible to mark how enzymes work in a graph?

The answer is Yes. We can show that with the Reaction coordinate diagram!

This is the Reaction Coordinate Diagram. And it shows how much energy we need/use to complete a reaction.

The X coordinate shows us the process of the reaction and the Y coordinate shows the amount of energy that we need for the reaction to occur.

As you know A is the reactant and B is the product.

For the reaction to be spontaneous, A has a higher energy level then B does.

It might be nice if the reactions just go to A to B straight, 

but that's not the case in most reactions. 

Instead, they need to overcome an "Energy barrier" to be a different molecule. The highest point of the energy barrier is called the Transition state. (‡) 

The transition state is where the most amount of energy is used and when the molecule is the most unstable.

The Free energy of activation, or activation energy (ΔG‡ or EA) is the energy barrier from A --> B. And it's shown as the difference between A and the transition state. 

Lastly, the Standard free energy change (ΔG˚ or ERXN) is the net change in energy levels between A and B. And it's the energy that is released once the reaction is over. And Enzymes lower the reaction's activation energy to speed up the reaction.

There are two ways to represent how enzymes bind and react to substrates.

One is called the Lock/Key model and the other one is the induced fit model.

The Lock and Key model was introduced by Emil Fischer. It shows that the enzyme and the substrate binds together perfectly like puzzle pieces. 

The Induced fit model was shown by Daniel Koshland. 

And It is a modification of the Lock and key model. 

The induced fit model shows that the substrate and enzyme does not quite fit perfectly. So the amino acid (side chains) of the active site changes it's position depending on the substrate while not giving any harm to the enzyme itself. After the reaction is done, the active site returns into it's normal position.

The induced fit model is used more often because it shows better how the enzyme and substrate binds.

Sometimes, an Inhibitor can bind to the enzyme and change the active site.

If that happens, the original substrate cannot bind to the enzyme.

So the substrate competes with the Inhibitor for their own purposes.

We call this Competitive Inhibition.

In competitive inhibition, just like the name says, the substrate and the inhibitor "race" each other to go to the active site.

And whoever goes to the active site first gets the enzyme. 

So if the substrate got in first, the reaction will occur and the inhibitor cannot bind to the enzyme and vice versa.

But the inhibitor doesn't always go to the active site. 

It can go to the allosteric site as well too!

Note: The Allosteric site includes every place of the enzyme except for the active site.

When in a competitive inhibition but the inhibitor goes to the allosteric site, we call this the Allosteric competitive inhibition.

In allosteric competitive inhibition, the same thing happens just like the competitive inhibition, but the inhibitor goes to the allosteric site instead of the active site.

However, For Non competitive inhibition, the substrate and the inhibitor doesn't necessarily compete each other.

Instead, the substrate can bind the enzyme until the inhibitor sticks to the enzyme's allosteric site. (Active site if there's no substrate)

In other words, Both of them can bind, but if that's the case, nothing happens.

Sometimes, enzymes are named for their functions. 

An example is DNA Polymerase, which is named because it is a polymer that works on DNA. The suffix "~ase" is tagged along with enzyme names.

If you're familiar with Amino acids, you will know that we can classify amino acids like aromatic, aliphatic groups.

Just like that, we can also classify enzymes as well!

Number 1: Transferase.

Transferases transfer specific functional groups from one molecule to another.

                           A + BX --> AX + B

Where A and B are the molecules and X is a functional group.

A good example of a transferase is an enzyme called Peptidyl transferase.

What this enzyme does is to link amino acids that is derived from the TRNA in translation to a growing polypeptide chain. And it also makes sure that they don't jumble up as well.

For short, Ala - Gly - His + TRNA - Ser --> Ala - Gly - His - Ser + TRNA

Notice how the Serine residue moved from one molecule to another.

Number 2: Ligase.

Ligases Joins two molecules together into one.

                                A + B --> AB

Speaking of Ligases, doesn't it gives you the feeling of our old friend?

That's right! It's DNA Ligase!

As we all know, DNA Ligase Fills up the Okazaki Fragments of the lagging strand at DNA replication. In other words, DNA Ligase joins the two strands of DNA to a single strand.

Number 3: Oxidoreductase.

Long name, but if we look closely, the name has the word "Oxido" and "Reduct". By that, we can figure out that this enzyme uses Oxidation and Reduction!

And that's true. They do use OR to catalyze reactions.

As you may know, Oxidation is when a molecule loses electrons and Reduction is when a molecule gains electrons.

                          A + B: --> A: + B 

NOTE: ":" is a pair of electrons.

An example of a oxidoreductase is something called Lactate Dehydrogenase.

Which it was a pain to learn about because it was in cellular respiration.

Lactate Dehydrogenase (LDH) helps the oxidation/reduction when Pyruvate and NADH reacts to make Lactic acid and NAD+ in Lactic acid fermentation.

              Pyruvate + NADH --> Lactic acid + NAD+

Click to Learn about Cellular respiration/Lactic acid fermentation.

Click to Learn about LDH in Wikipedia.

Number 4: Isomerase.

Isomerases helps to convert a molecule to one of it's Isomers.

                                 A --> B

An enzyme called Phosphoglucose Isomerase catalyzes the reaction of glucose 6 phosphate to one of it's isomer, fructose 6 phosphate.

                    Glucose 6 - P --> Fructose 6 - P 

Number 5: Hydrolase.

Hydrolases Uses water(H2O) to cleave a molecule into two other molecules.

                             A + H20 --> B + C

You might know what hydrolysis is if you learned about Amino acids.

Hydrolysis is the reaction where we use H2O to cleave peptide bonds of a polypeptide chain. And in hydrolysis, there is an enzyme called Serine Hydrolase. (Or Serine Protease) 

This enzyme uses the nucleophilic amino acid Serine at the active site to catalyze hydrolysis.

                       Pro - Cys + H2O --> Pro + Cys

Number 6: Lyase.

Lyases are a bit special. These guys don't use hydrolysis nor oxidation to catalyze a reaction. Instead, what they do is to break bonds of a single molecule and then form a double bond or a ring structure to make it stable.

This type of reaction needs one molecule for the forward reaction, but two for the opposite reaction. 

                                 A --> B + C

For an example of lyase, there's an enzyme called Argininosuccinate lyase.

And it catalyzes the formation of the amino acid Arginine and the dicarboxylic acid Fumarate from the molecule Argininosuccinate.

There, ASL breaks N - C bond and makes a new double bond between the "Broken" molecules to make it stable.

                  Argininosuccinate --> Arginine + Fumarate

And.... That's the 6 types of enzymes shown today!

Sometimes, enzymes cannot do it's work all by itself alone. 

They need some "Helpers" to do their job properly.

These "Helpers" are non protein molecules. We call them Cofactors.

We can go further into Cofactors and we can see that there are three different types of Cofactors that work differently.   

The first are Co - enzymes. They are Organic(C based) cofactors that are loosely bound to enzymes. Co - enzymes carries molecules or electrons.

We can see that in the lactic acid fermentation process, lactate dehydrogenase uses NADH as a coenzyme.

Another type of a coenzyme is Coenzyme A. (CoA) Which carries acyl groups instead of electrons.

The next are Prosthetic groups, and these can be organic or inorganic. 

Unlike Coenzymes, which are loosely bound, Prosthetic groups are either tightly bound or even covalently bonded to the enzyme.  

Do you know about Hemoglobin? It's a tetramer protein that carries oxygen around our body. In hemoglobin, there is a molecule called heme.

Heme is a prosthetic group that consists a Fe2+ (Ferrous) ion in the center of an organic heterocyclic ring called a porphyrin. Which is made by four pyrrolic groups connected by Methine bridges. (=CH-)

The last ones are Metal Ions. And they are inorganic. 

These Metal ions are Iron, Magnesium, copper, etc.

One more thing to say - Have you ever heard about Apoenzymes and Holoenzymes? Apoenzymes are enzymes that needs a Cofactor but doesn't have one. Apoenzymes are inactive - So it doesn't do anything unless the cofactor is bound to the enzyme.

Holoenzymes on the other hand, are Apoenzymes that has a Cofactor on it's side. To make it short, Apoenzyme + Cofactor = Holoenzyme.

When environmental factors, like PH or temperature changes too much that the protein can't stand, they undergo denaturation. And we've already seen how proteins undergo denaturation in amino acids.

Enzymes work the same way as well too.

For example, in DNA replication, DNA polymerase has to "Deal" with the negatively charged phosphate groups on DNA.

To stabilize the charges, DNA polymerase binds to the cofactor Mg2+ ion with one of it's aspartate residues. Which is deprotonated at normal PH conditions.

At normal PH, the negatively charged Aspartate residue will attract the positively charged Mg2+ ion through electrostatic interactions.

By using that Mg2+ ion, the DNA polymerase can stabilize the charge and can do it's job properly.

However if the PH is changed to low PH, the aspartate residue will be protonated and become into aspartic acid.

If that happens, the aspartic acid can't hold to the Mg2+ ion, and it can't function it's job. Note that H+ is not a cofactor. And a lot of H+ makes it a very acidic environment. Which is bad in this situation.

The other one is Temperature.

You may know what happens to proteins when they are in a very high temperature. They vibrate violently, breaking all the bonds that consist secondary, tertiary and quaternary structures. 

Just like that, enzymes denature in temperature changes as well too.

That's it for now in about enzymes and it's functions.

In the next time, things will be a bit easy to understand If you learn some basic stuff about Kinetics! The next is all about Enzyme Kinetics you know!

Click this to Learn Kinetics!

                                  ~Credits Section~

All Images by Khan academy, Wikipedia, and Wikibooks.

I do not own any of these sites.

If you want to see where the images came from or learn more, go to 

Wikipedia: https://en.wikipedia.org/wiki/Main_Page

Khan academy - MCAT: https://www.khanacademy.org/test-prep/mcat

Khan academy - Biology: https://www.khanacademy.org/science/biology

Wikibooks: https://en.wikibooks.org/wiki/Main_Page

                                    2016. 2/15

                              By Jane Kim <해인이가>

Welcome back everyone! Long time no see!

You know, I had to come in this late and write the article.... Because I wanted to tell you about the many things I've learned! 

Amino acid's left/right hand, a fact that has PH within it, and the one you have been longing to wait for.... How does AA's turn into well - shaped proteins?

We will be learning those soon and see some others as well!

Alright, Let's go! Into this wonderfully small world!

The first thing I want you to know is something called "Fischer Projection".

Fischer Projection is a 2D model for a 3D structure.

In Fischer projection, there are two types of AA's - The L and D types!

As you can see in the picture above, the fischer projection shows amino acids in four spots. The Amino group(Amine, or NH2), the carboxylic acid group(COOH), the side chain(R), and the bonded Hydrogen atom.

The L and D amino acids are like our left and right hands. 

Try to overlap your hands together. Do they overlap them perfectly? 

The answer is NO, and just like our hands, L/D AA's also cannot overlap each other perfectly. So we can say that L/D AA's are Non - superimposable to each other. 

The difference in structure is that the NH2 is on the left side for the L - AA while the NH2 is on the right side for the D - AA.

This makes the L amino acid to act as part of the system of proteins, enzymes, and lots of others that has to do with living things.

But for D amino acids, however, are not yet well - known. So it's hard to tell.

(If I get some more info, I'll tell you about it. Ok?) 

Now, here is the time I can tell you about Iso - electric point and zwitterions!

If we take a glycine molecule and show it, you will normally have NH2 and COOH. However, there can be a slight change to the structure. That is, when the COOH "knocks off" or donate the H+ to the NH2, making it into COO- and NH3+. This new glycine molecule has both positive and negative charge, while having a net charge of 0.(1 - 1) 

Like form B, we call this a Zwitterion!

Now, let me do a quick review of the difference between PH, PKa, and Ka.

Ka is the strength of an acid, or the "Degree of acid dissociation". 

PKa is the negative log of Ka, so the stronger the acid, the larger the Ka, the smaller the PKa, and vice versa.

PH is the measure(amount) of H+ in the solution. It is described as the negative log of [H+] ---> (The concentration of H+).

If you go back to the picture, you can see that the zwitterion can change it's 'form' depending on the value of PKa.

For form A, it has a PKa of 2.34, which means it's in a acidic solution.

This also tells us that the concentration of H+ has increased as well too.

Because a lot of H+ is in the area, the COO- can form a bond with the H+, making it back to COOH. And the net charge result is 1. (0 + 1)

On the other hand, for form C, it has a PKa of 9.6. 

This means it's in a basic - solution, and has a high PH which means that the concentration of OH- has increased.

Because of the OH- in the solution, the NH3+ can give it's H+ and become into NH2 again. So it's net charge is -1. (0 - 1)

Here's the fun part, you can now know at what PH can this glycine molecule become into a zwitterion! 

How do we do it? By using something called a Iso - electric point!

The Iso - electric point, also named PI for short, and this is the PH point where the AA has a net charge of 0.

How do we find PI? By calculating the total of our PKa's!

PI = (PKa1 + PKa2) / 2

With this equation, let's find the PI of glycine!

The equation is like this: (2.34 + 9.6) / 2 = 11.94 / 2 = 5.97

So we can see that 5.97 is our PI! Yay!

Oh, and one more thing. Different side chains(Acidic/basic) can change the PKa's to be different. And that's why all of the AA's PI's are different.

As we all know, we have 21 Amino acids in nature. 

Then, how do we sort these individual AAs in groups?

There's a lot of ways to classify Amino acids, but in this case, i'm gonna use it as this way:     

Amino acids --> Hydrophobic / Hydrophilic

Hydrophobic: Aliphatic group, Aromatic group, cyclic group.

Hydrophilic: Amide, S or Se containing group(neutral group), acidic, and basic group.  

Hydrophobic AAs are non - polar and does not like to form a bond with water while Hydrophilic AAs are polar and likes to bond and hang out with water.

We can split hydrophobic AAs to Aliphatic and aromatic groups. 

Both Aliphatic and aromatic groups has side chains are made out of Carbon and Hydrogen. the difference is - Aromatic groups has a cyclic hexagon - like structure(ex: benzene) while aliphatic groups doesn't.

Aliphatic groups includes Amino acids Glycine, Alanine, Valine, Methionine, Leucine, Isoleucine. Even though Methionine does have a sulfur atom, it is bonded to two Carbon atoms instead of an R - S - H bond. Making it nonreactive and non polar.

Aromatic groups includes AAs Phenylalanine, Tryptophan, Histidine ,and Tyrosine(Non polar because of it's benzene).

The Cyclic group includes one Amino acid - Proline. (Because of its odd, cyclic structure)

For the Neutral - or Hydroxyl, Alcohol, Amide, Sulfur, Selenium containing group, is in the hydrophilic group. Neutral groups includes Serine, Threonine, Asparagine, Glutamine, Cysteine(Cystine), and Tyrosine(slightly polar because of it's OH). 

Acidic groups have carboxylic acid(COOH) on their side chain, which is very acidic(strong hydrogen donor). Acidic members are Aspartic acid and Glutamic acid. In their deprotonated state, they become Aspartate and Glutamate.

Basic groups contain Amine groups as their side chain, which is basic.

Basic groups are amino acids Histidine(It's side chain is basic and aromatic), Lysine, Arginine.

These are the ways of classifying AAs in my way.

Say, after listening to all this amino acid stuff, doesn't it make you wonder?

About how does Amino acids turn and become proteins.

As it turns out, they go into a four step process for their well done protein structure - Primary, Secondary, Tertiary, and Quaternary!

Primary structure is about everything we learned - Amino acids, linked in peptide bonds. The very basics of proteins lives in here.

Secondary structure is where these primary structures folds themselves into Alpha helix(α helix) and Beta sheets.(β sheet)

An Alpha helix gets folded in a way that the C = O bond of the "nth" amino acid can form a Hydrogen bond with the N - H of the "n + 4 th" amino acid.

Which makes the Hydrogen bonds, drawn in dotted lines in the structure above.

For Beta sheets, however, they form a Hydrogen bond with the amino acid that's on the different strand (or "team") rather then the amino acid that's on the same strand.

Each of the chain of AAs in beta sheets are called Beta strands.

There are two kinds of Beta sheets - Parallel and Antiparallel.

Parallel Beta sheets has beta strands that starts and ends on the same side, while Antiparallel Beta sheets has beta strands that starts at a different side.

Just like on the Beta sheet picture above.

Now, what happens if a beta sheet is on 3D?

The Beta strands will be shown as arrows in the direction Nterm -> Cterm

and each Beta sheets will be shown as arrows more or equal to two.

The picture above is one of the beta sheets in 3D. Kinda looks like a hairpin, right? Well, just like that, the structure's name is called a Beta hairpin!

This Beta hairpin occurs when two antiparallel beta strands are together.

Do you see the paperclip - like loop in the 3D picture? That's called a Beta turn. It's made out of 2 ~ 5 amino acids and mostly consists Proline or Glycine.

These loops are important to keep the beta strands to be holded together and even to link with alpha helices. If they fold more, then what forms is called the Tertiary structure.

In Tertiary structure, there are two kinds - Globular and Fibrous. We are going to learn about Globular Tertiary structure. (Because I did not learn deeply to Fibrous Tertiary structures yet.)

Starting from Tertiary structure, you can call the whole thing a protein from now on. Globular proteins fold themselves as a ball shape in order to let it's hydrophobic stuff to the inside and the hydrophilic stuff on the outside.

This is called Hydrophobic interactions.

As well as the Hydrogen bond from the secondary structure, there are other bonds in tertiary structure to keep it stable.

This include Ionic bonds, and Disulfide bridges.(which is pretty hard to make, since they only happen in oxidizing environments) 

Quaternary structure are multiples of these tertiary proteins that are together.

Quaternary structures use the bonds that stabilizes tertiary structures.

An Example of a Globular quaternary structure is Hemoglobin. Which is a four - protein quaternary structure.

In quaternary structures, each protein that makes the quaternary structure is called a Subunit. When two of them are together, we call it a dimer.

For three of them, it's a trimer. Four is tetramer. And more then that are all called multimers. Each with their own name.

Normal proteins are formed with the steps above. But sometimes there can be an error in the step that causes the protein to be "Denatured".

Denaturation occurs if the protein gets enough "stress" to make it's structures to be broken and lowered down to the previous structure.

One thing to note though, Denaturation does not effect the Primary structure.

Now here are the Top 3 ways to Denature a protein:

Number 1: Heat.

Heat makes a lot of (Kinetic) energy that is strong enough to make the molecules vibrate rapidly. This breaks the Hydrogen bond of the secondary, tertiary, quaternary structure.

Number 2: Acid/Bases.

If an acid makes the solution to go below the PI, the concentration of H+ will increase. Making the negatively charged anion to be neutral.(Same with bases) Breaking the ionic bond of the tertiary, quaternary structure.

Number 3: Alcohols.

Alcohols like ethanol(CH3CH2OH), has this OH(Hydroxyl group) that can break the old hydrogen bond and form a new one. Making the secondary, tertiary, and quaternary structure to be denatured.

This is it, The end of this article!

The next one will be all about Enzymes.

Happy new year everyone, and happy studying!

NOTE: All pictures from Khan academy and Wikipedia. 

I do not own this picture nor site.

If you want more information, go to this link: 

https://www.khanacademy.org/test-prep/mcat/biomolecules

https://en.wikipedia.org/wiki/Main_Page

                                    2016 1/05

                                    By Jane Kim

                                    <해인이가>

Alright! I'm in a spectacular feeling about this biology article!

This time, we are going to learn about the building blocks of proteins,

Amino Acids!

21 of these little Amino acids all have different things that it can do - 

And we, are going study the basics of this adventure!

First of all, I would like you to know where amino acids are even a part of the body. Let's give you an example of a protein called "Hemoglobin".

Hemoglobin is a protein that's inside the red blood cells, and it's job is to pick up oxygens and bring it to the tissues(cells) all over the body.

The tissues uses the oxygen to generate ATP, which makes energy in order to live. And Amino acids are the ones that makes this important hemoglobin.

Here is the structure of a amino acid:

Amino acids are made by containing amine(NH2) and a carboxylic acid(-COOH) and a R and H.

Can you see the letter R? That's the side chain of the amino acid.

It's difference is what makes the amino acids all different.

Also, if you learned chemistry, you would know that there is a Carbon in the center part of the others. 

The carbon that is bonded with the NH2, R, and another carbon (alongside with a H atom that is not seen) is called the (chiral) alpha carbon.

"Chiral" means a carbon atom bonded to four unique other elements.

And the other carbon that's connected to the O, OH, and the alpha carbon is the carbonyl carbon.

Now, in order for these individual amino acids to become into a protein, The AAs should be linked together. A bond of two amino acids are called a "Peptide bond".

As you can see in the picture above, the first step is to "pluck" off the H from the N and combine it with OH-.

With the formation of a water molecule, connect the NH+ and the carbonyl C. Then a bond will show between the carbonyl carbon and the nitrogen. That's a peptide bond. And the molecule that's been formed is called a Dipeptide.

When these AAs join together, a polypeptide chain is formed.

For simplicity, We only write the N, alpha C and carbonyl C.

And so the polypeptide chain is like this: N-C-C-N-C-C-N-C-C

The side that the N is on is called the N terminal.

And the side that the carbonyl C is the C terminal.

We can imagine that the N-C-C to act as a "group" and call it a residue.

Just like AAs can make peptide bonds, you can also break down the peptide bond as well. in order to do that, we need to use a way called Hydrolysis.

There are two ways of Hydrolysis:

One is called "Acid Hydrolysis". And just like the name, we use strong acids with heat to break the peptide bonds.

The other one is called "Proteolysis". And for that, we use a special enzyme called protease. Unlike acid hydrolysis, proteolysis can let you pick which peptide bond to break.

For example, a protease called Trypsin cuts the C term of arginine and lysine. So if we have a polypeptide chain like this:

***- ^^^- arginine - @@@- !!! - lysine - &&& And input trypsin inside,

we would have it like this: ***- ^^^- arginine   @@@- !!! - lysine  &&&

Using proteolysis has a big advantage for experiments because you can cut wherever you want for the research. Pretty cool, right?

Now, we will talk about our most special Amino acids of them all - The AA show!

It looks like our contestants are Histidine, Proline, Glycine, and Cysteine.

What could possibly make them special?  

Well, let's look at our first stop, Histidine!

The thing that makes Histidine special is that it's pKa is approximately 6.5 which is close to physiological PH.

Now you might say "Umm, ok, the pka and ph is close.. So what?"

According to the Henderson - Hasselbalch equation, the pKa = PH + log [HA] / [A-].

If we make pKa > PH, or making the solution more acidic, the log [HA] / [A-] > 0. Which means that [HA] > [A-]. Meaning that a stronger acid will cause the formation of [HA], the protonated form.

If we make pKa < PH, AKA making the solution more basic, the log [HA] / [A-] < 0. So [HA] < [A-], meaning a stronger base will cause the formation of [A-], the deprotonated form.

Look here for more: 

http://www.mhhe.com/physsci/chemistry/carey5e/Ch27/ch27-1-3.html

For Histidine, since it's pKa is close to physiological PH, it can switch from protonated to deprotonated from and vice versa with just small changes of PH. Making it to a H+ donor/acceptor without wasting energy and Enable things like transporting protons to one molecule to another.  

Next is Proline and Glycine, the alpha(α) helix breakers!

Let's look at Proline first.

Unlike all the other Amino acids, you can see that proline has its R group to have a cyclic structure that has a bond with the amine(NH2).

So we say that Proline has a secondary alpha amino group.

Because of it's odd side chain, proline is very rigid then others. So it doesn't have a free rotation alongside the other atoms.

On the other hand, Glycine is the opposite of proline's structure.

Glycine's side chain is DA most simplest side chain in the HISTORY of side chains! It only has a H for it's side chain!

Because of it's simple side chain, it's very flexible then others.

But there is a catch for that simple side chain as well.

Remember the part where I told you about the "Chiral" alpha carbon?

Like I said, "Chiral" means four different elements bonded into a C.

But for Glycine, since it's side chain is a H atom, we have a double H for the alpha carbon. So we cannot say that the carbon is chiral for glycine.

Now, about this time now, I think you should know what an alpha helix is.

An alpha helix is a coiled up polypeptide chain. It kinda looks like this:

Even though Proline and Glycine has a role of forming alpha helixes, they introduces a Beta turn to the alpha helix.

A beta turn is a bend(or turn) in the primary structure.

For Proline, because of it's odd secondary alpha amino group, it makes a bend when making peptide bonds. Glycine also makes bends because of it's flexibility,

Our last Amino acid is Cysteine!

And so, what cysteine has is this little trick on it's sleeve to form a disulfide bridge when each of them are close to each other as a polypeptide chain.

It would be good to learn redox reactions before this.

https://www.khanacademy.org/science/chemistry/oxidation-reduction/redox-oxidation-reduction/v/introduction-to-oxidation-and-reduction

Now, you can see that Cysteine has a thiol group( C - SH) as it's side chain.

The Cysteine molecules are in a reduced environment. So if we put this in a oxidizing environment, the side chains will lose an H and the S will form a bond.

We call that bond a Disulfide bridge.

And I should really say - Cysteine has name issues!

Some are asking is "Cysteine" or "Cystine" the correct way to spell the name.

And the thing is, both of them are right.

The name "Cysteine" is used for it's usual form, where the disulfide bridge is not formed yet.

The name "Cystine" is used when it has a disulfide bridge, in an oxidized state.

So, you can think about it like this:

1, Cysteine is used for it's usual form, and inside it's reduced - intracellular space.

2, Cystine is used for it's bonded form, and inside it's oxidized - extracellular space.

And with that, the Amino acids show will end.

Amino acids are pretty fun, right?

I asked a question to my dad about why Amino acids are left out of the central dogma. (Because I was curious)

And the answer was simple - Since Amino acids makes proteins, we can say that they are the smallest protein unit. So it's not "left out".

Well, that's it for today's article. And I hope to see you again in the next one about Amino acids - DA protein makers! Swag! 

                                      2015, 12/30   

                                   Jane Kim (김해인)

                                                                                                                                                           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
<<해인이가>>

I'm in a desert, and I can choose either a compass or a map.

What would you choose if you were me?

I'd choose a map. And not just a regular map, i'm choosing a gene map!

And what I would like to show you this time is to know how to make this gene map when you have 3 character traits, along learning what is a gene order, and something called interference and COC.(And of course, our tester is a fruit fly)

Challenge accepted, let's do this!

First, I would like to show you a few stuff before we do anything.

We will use not 1, not 2, but 3 characters for this example.

And these characters are called Vermilion, crossveinless, and cut genes.

Vermilion: Fruit flies who has this gene has a brighter red eye color while the wild types has a dark red eye. And it's a recessive mutant.

Crossveinless: Most fruit flies that we know has a horizontal line and a bridge like vertical line on it's wings. However, crossveinless fruit flies does not have the "Vertical" lines on it's wings, and it's also recessive.

Cut: Normal fruit flies has round and smooth wings, but flies who has this character has it's wings that looks like it has been kinda shattered like glass or snipped by scissors. Again, it's a recessive mutant.

Now, look at the pictures with black backgrounds. There will be 3 of them.

If I write that in a simple form, it's just like this:

V, CV+, CT+ = 580       V+, CV, CT = 592      V, CV, CT+ = 45               V+, CV+, CT = 40        V, CV, CT = 89        V+, CV+, CT+ = 94 

V, CV+, CT = 3           V+, CV, CT+ = 5      Total progeny: 1448

NOTE: The genes that has + with them is the wild type, and the ones without the + is recessive mutant genotype.

Say, do you remember when we first learned about the term Centimorgans?

We normally write it as cM or m.u(map units). 

Well, this time might be the perfect time in order to use them in real life!

Sometimes you'd might want to know exactly how far is the distance between a gene loci from another, and want to know what gene is in where(To be specific, it's "what order"). A gene map can give you that much information about the genes in that way. It kinda looks like this:

                A                    34cM(mu)         B  12cM(mu)  C

   ---------|-----------------------------|-----------|--------

                |--------------------46cM(mu)-------------|

To make this, we have to know the distance between the three genes in order to make a gene map. 

To do that, we need to know something called a "gene order".

Then you might say "Ok, isn't the gene order V, CV, CT? All of the genotypes you showed me just looks like that!"

Well..... That could look like that at first, but you should know that there is a catch from the double mutant genotyped chromosomes.

Double crossovers moves their middle genes to it's sister chromatid, so we have to know the actual gene order. And we have to use that gene order once we have found it out.

How to find the gene order is easier then you would think. 

First, you need to find the parental genotypes and the double crossovered(DC) genotypes. 

Second, compare them and find the odd one.

Third, put the odd one in the middle, then put the rest of the genes where it was. 

And fourth, With the newly found gene order, find the distance between each gene loci. Sounds kinda hard, right? Follow the steps and you'll see.

Step 1: In order to find the parental genotype and the DC genotype, we just need to see what is the top 1 and 2 highest and lowest number of fruit fly progenies. The highest ones are the parental genotypes, and the lowest are the DC genotypes. In this example, we can now know that the genotypes V, CV+, CT+ and V+, CV, CT are the parental genes while the genotypes V, CV+, CT and V+, CV, CT+ are the DC recombinant genes.

Step 2: We have to compare the parental and DC genotypes together. 

That means we have to compare V, CV+, CT+  and V, CV+, CT together and also compare V+, CV, CT and V+, CV, CT+ together as well. 

We can directly see that the CT gene is the odd one out, therefore we put it in the middle and putting the other genes back in it's original location.

So we get V, CT, CV as our gene order, which means that the CV gene and CT genes swaps places.

Now, since we've figured out the gene order, the only thing left to solve is the distance between the gene locis! The only thing is, how?

Well first, let's arrange the stuff that we have just earned. 

If we put that in a gene map, it would look something like this:

                V                   ?cM(mu)           CT  ?cM(mu) CV

   ---------|-----------------------------|-----------|--------

                |--------------------?cM(mu)--------------|

Remember in the 20th line where I have wrote all the possible progenies?

As you all have learned that the homologous chromosome pairs has to undergo "Genetic Recombination" for other progenies to appear in their phenotypes, they can do the same thing! A easy way to figure out the map units is to figure out where the crossing over happens, what kind of genotypes can be formed, and make that into a percent! 

According to the gene map above and the total genotype progenies, we can see that the distance between V and CT or CT and CV can undergo genetic recombination.  

                V                   ?cM(mu)          CT+  ?cM(mu) CV+

   ---------|-----------------------------|-----------|--------

                                        X                         X

               V+                   ?cM(mu)          CT  ?cM(mu)  CV

   ---------|-----------------------------|-----------|--------  Hmmmm...... What expected genotypes can occur when crossover from the left happens? Right! We can get V, CT, CV and V+, CT+, CV+

Then what about the recombination on the right? 

We would get a result of V, CT+, CV and V+, CT, CV+

And what if both happened? A result of V, CT, CV+ and V+, CT+, CV would occur, and the other two left are the parental genes.

To figure out the map units, we need to add up all the genotype's progeny numbers, divide that to the total progeny number(Which is 1448), and multiply it by 100. 

So for the V ~ CT range, we just need to know what are the progeny numbers from each genotypes that can occur from that range's genetic recombination.

But there is a catch - I just noticed this. You were probably just thinking to add the progeny numbers of the genotypes V, CT, CV and V+, CT+, CV+(89, 94) and solve the question, right?

Well, that is wrong! Think about the genotypes that has undergone a double crossover, that kind of genotype does both of the recombination V ~ CT and CT ~ CV. So you should put their progeny numbers as well.

Then we can know that the distance between V ~ CT is: (89 + 94 + 3 + 5) / 1448 = ~0.132 

0.132 * 100 = 13.2%!

CT ~ CV = (45 + 40 + 3 + 5) / 1448 = 0.064

0.064 * 100 = 6.4%

Then what is the distance between V and CV? 

We just have to add the two distances together!

13.2 + 6.4 = 19.6

Which means our gene map looks like.......!

                V                   13.2cM(mu)       CT  6.4cM(mu)   CV

   ---------|-----------------------------|--------------|-----

                |--------------------19.6cM(mu)--------------|

Yeah! It's all done! The gene map is finished!

But, even with all this excitement, we still have one more mission left.

I would like you to learn a bit about probability for this one, assuming that you have, here it is.

Let's just say that we don't know the progeny number of the double recombinant genotypes. If that's so, then we would figure out the number with this "algorithm". Like below. 

0.132 * 0.064 = 0.0084 

0.0084 * 1448 = 12

(We did this based by the progenies of each recombination)

We got 12 as our answer, right? 

Then count the total number of progenies from this chart:

V, CV+, CT+ = 580       V+, CV, CT = 592      V, CV, CT+ = 45               V+, CV+, CT = 40        V, CV, CT = 89        V+, CV+, CT+ = 94 

V, CV+, CT = 3           V+, CV, CT+ = 5      Total progeny: 1448

Since V, CV+, CT  and V+, CV, CT+ is the double recombinant genotypes, we would get our answer as 3 + 5 = 8

Now what just happened here? We thought the total number of progenies was 12, when we actually got 8!

All of this is because of Interference.

When meiosis happens and genetic recombination occurs, if a gene loci(Or "spot) has done genetic recombination, it decreases the probability that the gene loci next to it will also undergo recombination.

The interference tells us how strong a crossover from one DNA gene loci place interferes with the others.

To figure out the interference, we need to figure out something called COC, and it's short for "Coefficient of coincidence".

How to find the interference is to follow this formula: 1 - COC

COC = Observed # of DC progenies / Expected # of DC progenies.

Luckily, we have already figured out the COC value, which is 8 / 12 = 0.66

Then the Interference value is 1 - 0.66 = ~0.33

And that's why the observed & expected progenies are different.

So a long talk! But we finally figured out what is the gene map for this fruit fly and about interference!

Oh, and guess what? Here are the links to some of the helpful sites that will help you clear any more confusions!

https://en.wikipedia.org/wiki/Coefficient_of_coincidence

https://www.youtube.com/watch?v=J-aTSriYnak

https://www.ndsu.edu/pubweb/~mcclean/plsc431/linkage/linkage3.htm

I will be writing a another article when I learn some other interesting stuff!   

Thanks for reading this!         

                                By Jane Kim    <<해인이가>>

update: 2015-10-12

Replication!

The source of how we grow, and how we get healed, and how we are even can be twins!

Yes, just like the examples I showed you, even tiny tiny cells in your body replicates and even the DNA inside of it!

Technically, if a DNA is replicating, it means something good might happen!
And this, my friends. Is the stuff we will learn about in today.
The "Advanced source" of DNA replication. Let's get ourselves an egg roll, shall we?

The first thing you know about DNA is the writing I showed you in the last one, whereas the DNA has a part going in the 5` to 3` direction, and the 3` to 5` direction.
DNA has it's deoxyribose backbone and it's four nitrogenous pairs - Adenine, Thymine, Cytosine, Guanine. Because of that, you can easily tell which base pairs are going to be with which base pairs.
There are 8 basic enzymes in order to start you DNA's replication, and those are - Helicase, SSB's, Topoisomerases, RNA primase, DNA polymerase, Sliding DNA clamps, RNAse H, DNA ligase.
And below are the jobs that these enzymes do.

Helicase: This enzyme uses ATP hydrolysis energies to break apart the DNA strands. Their limit is the replication fork, which tells the enzymes where to replicate and where to leave it be.

SSB("Single Strand Binding" proteins): it's job is to make sure that the opened up strands of DNA's doesn't get twisted back to it's normal shape.

Topoisomerases: It makes sure that the other DNA's doesn't get unwounded or get to coiled with each other(Supercoils). This enzyme also uses ATP for it's job, like the Helicase.

RNA primase: Before the other enzymes could start putting in some DNA, it has to have a starting line first, and that 'starting line' is the RNA parts by the RNA primase. Without it, the other enzymes can't do their work of putting the DNA parts.

DNA polymerase and the Sliding DNA clamp duo: Those two are excellent combos. DNA polymerase puts the DNA parts after the 'starting line RNA' that RNA primase has left, and the Sliding DNA clamp makes sure that the DNA polymerase doesn't get side - tracked.

RNAse H: This guy over here removes the RNA parts left by RNA primase.

The second DNA polymerase: Like the first DNA polymerase, this enzyme places DNA blocks in the parts where the RNAse H has removed the RNA parts. It makes Okazaki Fragments in the process, which is a gap between the DNA parts from the RNA places and the original DNA parts.

DNA ligase: This little enzyme fills out the gaps of the Okazaki Fragments, and finishes the replication!

For beginners, teachers shows these enzymes as if they are coming and going on their own, for the sake of simplicity. However, they don't actually just come and go out of nowhere! They are all doing their work based on the structure of the main Holoenzyme!
Look at the 3rd picture which looks like an ant's head with some sort of green stuff stuck on it's antenna, did you got that picture?
Now, I will show you what is the Holoenzyme and what's it's structure.

DNA clamp holder: That's the place where the holoenzyme holds the DNA clamp and uses it and reuses it over and over again.

γ complex(clamp holder): When replication starts, it makes the DNA clamps and holds it in the clamp holder.

τ protein: This protein bundles to hold the DNA polymerases.

That's about it, that's the holoenzyme, and the replication picture is the picture below the holoenzyme picture.
Oh, and one more thing! remember this: 5` to 3` direction is where the enzymes go, so the DNA's have a leading strand and the lagging strand.
The leading strand has no problems for replicating, because it's strand is in the 5` to 3` direction. But the lagging strand is a 3` to 5` direction, so you have to do stuff all over and over again, making a lot of Okazaki Fragments.

Now, with all of this replication stuff done, do you have any questions?
If not, then you could go for a dessert break. And if you have one, you are a wise student! Ok, what's the question? I'll give you one!
Have you ever thought what was the actual stuff that the DNA polymerase places down after the RNA parts? Sure, it's DNA, but what does the actual backbone looks like, and the actual structure, and it's molecule! What are those kinds of stuff, anyways?
And here is the answer - Deoxyribonucleotide triphosphate
This thing - When you look at it, if kinda looks familiar. That's right! it kinda looks like the structure of a DNA nucleotide!
The thing is, instead of having one phosphate group connected with the 5` end, this DTP(Deoxyribonucleotide triphosphate) has 3 phosphate groups!
And I'll tell you why.

DTP is like the DNA nucleotide with 3 phosphate groups instead of one.
Those 3 phosphate groups have a name - The α group phosphate, β group phosphate, and the γ group phosphate.
Whenever replication occurs, the DNA polymerase places this molecule one by one. But here is the thing, whenever the DTP lands on a another DTP, guess what happens to the triphosphate groups?
The β group phosphate, and the γ group phosphate separates from the molecule!
Which gives us a result of the normal DNA nucleotide and 2 individual phosphate molecules!

Well, that's all about replication that I have learned today.
Already on my 3rd page of this activity! What would the 4th one be?
Well, better go and find the flour of videos soon!
See you at next lesson!

By Jane Kim
<<해인이가>>

by jane kim(김해인) 2015.10.11에 작성: Vector개념을 배우고 이를 코드로 작성해 보았습니다..

코드 실행: https://www.khanacademy.org/computer-programming/pvector-instance/6479887750266880

/**

Try to catch the ball! It's harder when you move, but easy when you stay..... :)

Line 37: The reason why I used constrain() is because sometimes(or mostly) when you move your mouse in further directions, the ball can get off the canvas. Making the user a bit fustrated.

So, I just used constrain() to command the code where the ball can go without further ado.(it can only move inside the canvas!)

*/

background(97, 237, 255);

//I'm puting this in a function just in case somebody tweaks this code and gets an error. (Nobody wants that to happen!)

var ball = function(x, y) {

    translate(x, y);

    noStroke();

    fill(255, 0, 0);

    ellipse(0, 0, 30, 30);

    fill(255, 255, 255);

    arc(0, 0, 30, 30, 0, 180);

    fill(0, 0, 0);

    ellipse(0, 0, 5, 5);

};

mouseMoved = function() {

    background(97, 237, 255);

    //Creating two new PVectors, mouse and center.

    /* The mouse is a vector showing us where the mouse is, and the center is just the center of the canvas.(In vector scale!) */

    var mouse = new PVector(mouseX, mouseY);

    var center = new PVector(width/2, height/2);

    /* The mouse.sub(center) is in order to let the ball to fallow the mouse. Without it, it will go away twice from the position the mouse is currently on.

NOTE: mouse.mult(3) is to make distances from the mouse and the ball 3 times further.*/

    mouse.sub(center);

    mouse.mult(3);

    //Since translate() is not with either pushMatrix() or popMatrix(), we have to reset the code by using resetMatrix().

    resetMatrix();

    translate(width/2, height/2);

    ball(constrain(mouse.x, -185, 185), constrain(mouse.y, -185, 185));

};

/**

Note: This is just a practice program for my vector skills, not for khantober!

*/

by jane kim