The gene map and interference

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>>