Branch Diagrams to Solve Phenotype and Genotype Questions

A lot of students approached me about question 2.27 from the Sanders and Bowman textbook.  I can't use their question directly, but I came up with a similar one.   I like this question because it can be used to practice how branch diagrams are used, and for both phenotype and genotype.

Try this out on your own.  I have two videos for the answers.   One is for a-c which are phenotype-related and were easier for most students.  The second video is for d&e which involve genotypes.  Note that you can solve these without using branch diagrams!  The student Study Guide and Solutions Manual shows other ways to correctly determine the answers.

Part 1:  Phenotypic ratios.

Part 2:  Genotypic ratios

See?  Branches aren't so hard!

"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:

Ratios, ratios, ratios

In classical genetics, phenotype and even genotype ratios are a mainstay for determining inheritance.  You can memorize the ratios, but it's important you understand what they're based on.

In this exercise, assume we're dealing with simple dominant/recessive relationships for two traits in a bean plant.  The bean can have yellow or orange pods and can be inflated or constricted.  Constricted or orange pods are novel (new) traits.  You cross a true-breeding orange, constricted bean line with one that is true-breeding for yellow, inflated pods.  The F₁ seeds are planted and all grow up to bear fruit which are yellow and inflated.  The F₁ are allowed to self-fertilize.
You have several tasks:

1. Design appropriate gene symbols.
2. Use a branch diagram to show the expected phenotypic distribution of the F₂ progeny.
3. Of the F₂ progeny, those with orange pods are selected and allowed to self.  What is the expected phenotypic distribution among these select F₃ progeny?
4. Demonstrate the expected ratios of progeny from a test cross for the original F₁ plants using a Punnett square.

Here's a full solution, but if you want to just see the solution for part 3 (as my students wanted) you can see a shorter version in the second YouTube video.

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Here's just the answer to 3: