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

*as*

__taught__*. 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.*

__learned__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:

- 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 (2
^{n}, where*n*=the number of traits). - 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. - 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*. - 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*." - ...and so on.

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:

The straightforward way (using the logic in Figure 5.8):

"Working backward from the formula" (most popular attempted technique, but almost nobody wrote down all the classes and strategically filled in the numbers):

Branch diagram way:

So … solvable if you're strategic!

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