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I have been learning about recombination frequencies, but an still getting a bit confused despite having gone over many of the links in Google regarding them. I was wondering if someone could verify whether the following is correct.

  • To calculate recombiation frequencies and thus create a genetic map from this, you have to have true breeding parents for two or more different genes.
  • You have to cross these parents, so that you have an F1 generation consisting only of heterozygotes
  • You then have to backcross the F1 individuals with either of the parents. If these genes are all located on different chromosomes you will just get independent chromosome segregation and therefore a 50/50 chance of each allele occuring with the other one in the progeny. Whereas if they occur on the same chromosome, then in the F1 heterozygotes the alleles from each parent will only segregate differently into the gametes if recombination occurs.
  • To calculate the recombination frequencies we therefore take the number of offspring for which the combination of these two alleles is different from that observed in the parents- we call these the recombinants.

Is this correct?

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This answer is mainly copy-pasted from my answers here and here.

The importance of polymorphism

To understand what generation to backcross, it is central that you understand why we need polymorphic loci.

If at least one locus is homozygous

If a recombination event happens between two loci where at least one is homozygous, then you would not see anything. Consider for example the following strand sequences in a diploid individual

-----A-----B----
-----a-----B----

Whether a recombination event occurs or not, the two possible chromosomes passed down to an offspring are

-----A-----B----
-----a-----B----

Therefore, you cannot tell whether recombination has occurred.

If both loci are heterozygous

Now consider the following individual

-----A-----B----
-----a-----b----

If no recombination event has occurred between the two loci of interest then the two possible chromosomes that will be passed down are

-----A-----B----
-----a-----b----

If on the other hand a recombination event has occurred between the two loci of interest then the two possible chromosomes that will be passed down are

-----A-----b----
-----a-----B----

Therefore, you can tell whether a recombination has occurred or not.

Statistics of recombination

There is a tiny bit of maths below. These equations are mainly for curiosity as one can understand the answer without understanding the maths behind it.

Definitions of $r$ and $M$

You are getting confused between two different statistics

  • The rate of recombination $r$ between two loci
    • $r$ is the probability for two sequences found at two loci to remain in the same gamete after recombination occurred. This probability cannot be greater than 0.5 ($0 \le r \le 0.5$).
  • The distance in Morgans $M$ (or more commonly in centiMorgans) between two loci
    • $M$ is the expected number of cross-over that occurs between the two loci.

Morgans and centiMorgans

You will note that I talk in Morgans rather than centimorgans which is unusual in the literature but it helps at conveying the intuition of what it means. if $M= 150$ centiMorgans $= 1.5$ Morgans, then the expected number of cross-over between the two loci is 1.5. Below are some more explanations on these two definitions with some drawing :)

Case study with loci A and B

While $M$ and $r$ are closely related, they are not exactly the same thing. Consider the following sequence with the loci A and B

---[A]------------------[B]---

Let's assumed the two loci are very far apart and $M=2$. The probability of having exactly $k$ crossovers is therefore given by a Poisson distribution with rate $M=2$

$$P(k) = \frac{e^{-M}M^k}{k!}$$

Let's say for a given case that $k=1$ (a single cross-over occurred). This cross over is represented by a "/" below

---[A]-------------/----[B]---

Here clearly the two sequences at loci A and B will be separated. Let's say now that $k=2$ (two cross-over occurred).

---[A]---/-----/--------[B]---

Here, even if crossovers have occurred, the two sequences at loci A and B will remain together. Only the sequence in between the two cross-overs will come from the homologous chromosome.

Relationship between $r$ and $M$ - in words

You might see it coming from the previous section. The probability $r$ of these two loci A and B to be separated via recombination is the probability of an odd number of recombination events to occur between them (knowing that $M$ is the expected number of cross-overs).

Relationship between $r$ and $M$ - in equation

Let's calculate first the probability $p_{even}$ that an even number of cross-overs occur. This probability is just

$$p_{even} = \sum_{k = 0}^{\infty}{e^{-M}M^{2k} \over (2k)!}$$

, where I just added the constant $2$ before $k$ at both the numerator and denominator. With some algebra and trig, one can show that

$$p_{even} = \sum_{k = 0}^{\infty}{e^{-M}M^{2k} \over (2k)!} = e^{-M}\sum_{k = 0}^{\infty}{M^{2k} \over (2k)!} = $$

$$e^{-M} \: cosh(m) = e^{-M}\left ( \frac{e^{M} + e^{-M}}{2}\right ) = {1 + e^{-2M}\over 2} $$


As, $r = p_{odd} = 1 - p_{even}$,

$$r = 1 - {1 + e^{-2M}\over 2} = {1 - e^{-2M}\over 2}$$

Here we go! We have our relationship between $r$ and $M$! Let's graph it

Relationship between $r$ and $M$ - on a graph

I just graphed the above equation in Mathematica (Plot[y = (1 - Exp[-2 M ])/2, {M, 0, 5}])

enter image description here

Here is the same graph but zoomed on lower values of $r$ and $M$ (Plot[y = (1 - Exp[-2 M ])/2, {M, 0, 0.1}])

enter image description here

We clearly see from the graph that for low values of $M$, as $M$ increases $r$ increase quasi linearly ($M≈r$). For greater values of $M$, $r$ still increases but slower and slower until reaching an asymptote / plateau at $\frac{1}{2}$. $r$ is indeed bounded between 0 (when $M=0$) and $\frac{1}{2}$ (when $M=\infty$).

Note, the fact that the sum of probabilities of every even $k$ in a Poisson distribution is always lower or equal to 0.5 is an interesting math fact in itself!

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  • $\begingroup$ Some of your math was incorrect, as mentioned [in a comment] in the other post that you referenced, so I corrected it. I kept it simple and straightforward, using only common trig identities of $cosh(x)$. $\endgroup$ – Charles Sep 20 '17 at 3:20
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Yes that's how it works as far as I know

Because the theory part is pretty straightforward.of the genes are on the Same chromosome, the chance of them being separated in segregation during meiosis is lower

Consider a simple example

---------A--b--------------c----d-e---------------------------------f

Let this be a rope with 6 point a b c d e f If I were to make a random cut anywhere here, the chance that a,b will get separated is much lower than a,f....because a and f are so far apart that any cut in the middle will ensure them to be separated. If u look at it again...the chance of d,e to be separated is EXTREMELY low

This is the principle of LINKAGE

Now let me say gene a stands for tall, b for blue and f for bushy...(don't judge the characters)

Now if this is a part of a chromosome on the parent, and meiosis is taking place We all know that crossing over too is a phenomenon involved here

Now during crossing over.....a and f are bound to get separated for sure, so they will be independently assorted, obeying the law of Mendel

But when we talk about a and b, they may or may not get separated in the various gametes

The closer they are, the lesser chance of them being separated and thereby, higher chance of them occurring in the same gamete and therefore, higher chance of them being expressed in the same offspring(assuming they are dominant)

lets assume we cross blue tall(dominant) with green dwarf(recessive) As you know f1 will be blue tall(no incomplete or co dominance please)

Now if I take this blue tall and cross it with parental green dwarf

You will have very high number of blue tall and green dwarf(remember, there are alleles of SAME GENE....AND OCCUPY SAME LOCI)(therefore.....high chance of both occurring together in the same gamete)

Therefore the number of recombinant you get will be very low(blue dwarf and green tall........as the chance of these characters entering different gametes is very low)

Therefore we can conclude that higher the number of recombination....higher the chance of the genes segregating which implies a greater distance between the genes)

Therefore a higher recombination frequency implies a higher distance between the 2 linked genes

I hope u have understood the concept and are not struggling to byheart it as a force Thanks

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