"Reverse Mapping" of loci

It's one thing to take a table of recombinant data and then plug the numbers into a calculator to get the map distances.  It's another to take a map and generate the data yourself.  When I started teaching, I was terrified of generating exam questions.  I wasn't sure how to create the data for the table.

In the textbook I chose for this course (Genetic Analysis:  an integrated approach by Sanders and Bowman - 1st edition) they have a really good figure I showed in class for how to derive those kinds of data.  Figure 5.8 (below) shows how to take a map and then break down the distances into sections for each crossover class.  I solved this in my class and asked if it made sense.  I got a lot of nodding heads, and nobody looked too terror-stricken.  So I put it on the exam.

Well that proved interesting!  I went over the exam with one of my genetics sections afterwards, and some people couldn't remember me having done it.  I recalled that I'd used the side board to solve it - something I don't normally do - and a lot of people remembered the instruction.  Not so much the content, but that there was a solution that wasn't very difficult.  This underscores an important fact about teaching genetics ... it's not so much taught as learned.  You can watch a proof and understand it, but you actually have to struggle through and do it yourself or the knowledge won't take hold.

Here's the figure in two parts, starting with Part (a):

 

This first part of the figure illustrates the basic concept that "map distance reflects the probability that a crossover will create a recombinant chromosome."  Two chromosomes, in fact:  one of each type in approximately equal quantities.  The map distance shown here says "10% of all gametes will be recombinant".  You probably remember that 10% is the same as 0.1, so my explanations will use either convention.  There are two ways we can see recombination (i.e. they go from coupling:  AB or ab, to repulsion:  Ab or aB).  Half (5%) of the recombinants will be Ab and the other half (5%) are aB.  How many gametes are parental?  100% (of the gametes) - 10% (recombinants) = 90%.  45% of those are AB, 45% are ab. 

Now look at three linked loci, as shown in Part (b):

We've added a layer of complexity:  double crossovers.  Note that the easy way to solve this is to go stepwise:

1. Find out how many phenotypic classes you have.  There should be two parentals (largest numbers of equal magnitude), two double crossovers (smallest numbers of equal magnitude), and two classes that fall in between, with each class consisting of two values approximately equal to each other.  I taught my students that we should expect 8 phenotypic classes for crosses of three traits (2ⁿ, where n=the number of traits).
2. Calculating the double crossovers uses the product rule.  In the example above, we expect a 10% chance of a crossover between a and b, and of those, 20% will also experience a crossover between b and c.  0.1 x 0.2 = 0.02.  Half of those (0.01 or 1%) will be AbC, the other will be aBc.  This is like the calculation you'd do for genetic interference in my other blog posts.
3. Now let's look at the parental classes.  The chance of NOT having a crossover between a and b is 90%  (1.0-0.1 =0.9).  Of those, 80% will also not cross over between b and c (1.0-0.2=0.8) so the total number of parentals is 0.9 x 0.8 = 72%.  36% will be ABC; 36% will be abc.
4. Apply the logic for aBC and Abc.  You can see the formula is now 1/2(0.1)(0.8) which is saying "10% of a crossover between a and b and 80% of NO crossover between b and c."
5. ...and so on.

Most students on the exam worked backward from the formula, which CAN work, and I'll show you how in a video below.  Here's the formula just to refresh your memory. 

Note there's also the ability to use a branch diagram, so I'll show that kind of solution as well.

So back to my philosophy:  to learn genetics, you have to DO genetics. With that theory in mind, try this one.  I'll solve it three ways in the solutions below the fold.

Click to reveal solutions:

Intragenic Recombination

In other parts of this blog you saw how genes can be mapped with respect to each other using a simple rule:  the close two genes to each other, the less likely there can be a recombination event between them.  This means that the proportion of crossovers is linked to gene distance, so Sturdevant, a student of T.H. Morgan, developed the first classical mapping technique - essentially the one used today.

What's the CLOSEST two loci can be to each other?  From a theoretical standpoint, it would be to the smallest unit that can be heritable.  Scientists knew these to be "genes", and in the 1950s (before the double-helix structure of DNA was known), the biologist Seymour Benzer wanted to ask what the smallest measurable distance of crossover would be.  Genes weren't understood the way we know them now - as linear arrays of nucleotides that portray information not unlike how certain chains of letters make up words and sentences.  Perhaps they were very complicated structures, and crossover couldn't occur within them?  Or is it possible that the nucleotide letters in a gene align up between homologous regions and crossover can occur?  Benzer used rII mutant viruses to answer the question.

rII mutants cannot infect E. coli strain K12.  They *can* infect E. coli strain B, but the plaques that form have unusual phenotypes.  At high density of plaque formation, though, they're hard to distinguish from each other.  In the course of investigating the genetic nature of the rII locus, Benzer found out there are two genes there, which he named A and B.  A functional A and a functional B gene product (i.e. protein) is required to infect strain K12.  If you coinfect bacteria with an A mutant and a B mutant, lots of progeny form.  However, if recombination can occur within a single gene, two A mutants (let's call them rIIa and rIIa') can - on rare occasions- create two recombinants (we'll call them rIIa⁺ and rIIa'').    The trick is to find out how many wild type (rIIa⁺) phage are created.  So Benzer's trick is simple:  find out your total progeny by counting how many phage result from a coinfection of E. coli B strain by plating it on, well, a B strain lawn.  You can find all the wild-type phage (rIIa⁺) by plating them on a K12 strain lawn, and use the Studevant mapping formula:  distance = #recombinants/total progeny x 100.

However, since you don't see ALL the recombinants - only the wild type rIIa⁺because rIIa'' look just like the parentals - you need to multiply the number of K12 plaques by 2.  For every wild-type phage, you'll make one of those new "double mutant" alleles (A bad name for the rIIa'' genotype, but somewhat descriptive.  Hopefully you're following the logic here!).

In the two videos below I go over these concepts.  The first is the theory I've given above.  The second shows how to use the formula.  I didn't point out that coinfection must occur in E. coli strain B, but you probably could figure that out for yourself!