Gene mapping involves both logic and numeracy and this can scare some students. Below are examples that work their way from very straightforward to more challenging.
Example: Alien crosses
Here's an example I used in class a little while ago. Lucky for us, the aliens exhibit Mendelian dominant/recessive characteristics! They're also diploid, and have linear chromosomes.
Here's the setup:
You discover an alien species that displays the same type of inheritance that is commonplace with diploid animals on Earth. You cross two true-breeding individuals and the F1 displays these traits: pear-shaped head, gray skin, and normal fingers. The F1 is testcrossed to give the following data:
Cross of aliens
|
||
91
|
pear-shaped head, glowing
fingertip, and green skin
|
|
6
|
pear-shaped head, glowing fingertip, and gray skin
|
|
1
|
pear-shaped head, normal
fingertip, and green skin
|
|
1
|
round head, glowing fingertip, and gray skin
|
|
7
|
round head, normal fingertip, and green skin
|
|
506
|
pear-shaped head, normal
fingertip, and gray skin
|
|
85
|
round head, normal fingertip, and gray skin
|
|
491
|
round head, glowing fingertip, and green skin
|
Define gene symbols and create a genetic map that shows how these traits are arranged on the chromosomes. Be sure to mathematically correct for double crossover events. Calculate interference.
Determining Gene Order
I often get asked about how to set up the F1 chromosomes in order to figure out the gene order. The textbook uses examples where all the wild-type alleles are on one chromosome and all the mutants are on the other. This is called a coupling arrangement:
e.g.
a+ b+ c+
==========
a b c
However, it's certainly permissible to have an F1 organism that has some alleles in repulsion:
e.g.
a+ b c
==========
a b+ c+
For the first case (all in coupling), if the het is derived from two true-breeding parentals, they might have the genotypes of:
a+ b+ c+ a b c
========== x =========
a+ b+ c+ a b c
The double crossover class from the testcross would be:
a+ b c+ a b+ c
========== or ==========
a b c a b c <=== This came from the testcross parentFor the second (some repulsion), the double crossover classes from the testcross would be:
a+ b+ c a b c+
========= or ==========
a b c a b c <=== This came from the testcross parent
Here's an exercise to help you with this concept.
A solution is shown below (click on the YouTube icon to go to the YouTube site so you can view it in full screen and high definition).Example: Worksheet with trickinessness built in
Okay, I know trickinessness isn't a word, but what I mean is that we'll start with a simple two-trait cross to warm you up, then go to three traits that are all arranged in coupling. We'll move on to other examples (not necessarily in this order) that have some traits in coupling and some in repulsion, missing phenotypic classes and also when the data demonstrate independent assortment of one locus!
You might find it really helpful to download this worksheet and print it out. I strongly encourage you to try these exercises out for yourself and only when you think you're done (or are really stuck!) should you look at the solutions video!
Go to http://tinyurl.com/F17classical for the worksheet.
Mapping two loci
Two true-breeding lizards were crossed. Two mutant traits were found in the parents: bent tail and curled claws. The F1 lizards were all wild-type in appearance. The F1 females were testcrossed, and the offspring were sorted to obtain these data:
Two true-breeding lizards were crossed. Two mutant traits were found in the parents: bent tail and curled claws. The F1 lizards were all wild-type in appearance. The F1 females were testcrossed, and the offspring were sorted to obtain these data:
Phenotype
|
Number
|
wild type
|
471
|
bent tail, curled
claws
|
504
|
bent tail
|
11
|
curled claws
|
14
|
- Define appropriate gene and allele symbols according to standard conventions.
- Map the distance between the two genes.
The solution for the question is below.
Adding a locus to make it three!
Now let's take a look at what happens when we add another gene! This is question 1b.
*don't forget to try out the worksheet first!*
Two more true-breeding lizards were crossed. Three mutant traits were found in the parents: bent tail, missing thumb, and curled claws. The F1 lizards were all wild-type in appearance. The F1 females were testcrossed, and the offspring were sorted to obtain these data:
Phenotype
|
Number
|
wild type
|
182
|
bent tail,
missing thumb, curled claws
|
176
|
bent tail,
missing thumb
|
5
|
missing thumb,
curled claws
|
52
|
bent tail
|
55
|
bent tail,
curled claws
|
2261
|
missing thumb
|
2279
|
- Diagram the arrangement of alleles on the two homologous chromosomes for both parents (P generation) and the F1.
- Draw a genetic map based on these data. Be sure to mathematically correct for double-crossovers.
- Calculate interference. Explain what this value means
The solution for the question is below.
(You can view a full-screen version by clicking on the YouTube logo at the bottom-right corner of the video above).
What? What? You aren't naming the "normal" traits?
This (as well as the previous) question can confuse some students because it conforms to the convention of not saying what the wild-type features look like. Only mutations are written in the description. If a feature isn't noted, then you can assume it's wild-type. Also be careful when determining the order of the genes and whether they're in coupling or not! It's from a worksheet that is available online
A homozygous female with mutations for vestigial wings, blue
body, and purple eye colour was mated to a wild-type male. All the triple hets had blue bodies. The F1 females were testcrossed
and eight classes of progeny were classified as follows:
Phenotype
|
Number
|
vestigial, blue, purple
|
1572
|
wild-type
|
1553
|
blue, purple
|
129
|
vestigial
|
118
|
blue
|
29
|
vestigial, purple
|
39
|
vestigial, blue
|
1
|
purple
|
4
|
- Define appropriate gene and allele symbols according to standard conventions.
- Diagram the arrangement of alleles on the two homologous chromosomes for both parents (P generation) and the F1.
- Draw a genetic map based on these data. Be sure to mathematically correct for double-crossovers.
(you can get a full-screen version by clicking on the YouTube logo at the bottom-right hand side of the video above).
One kind of trickinessness
I threw in a different twist for this one. The worksheet is available online. You can use the same strategies as for the other two questions I modeled, but you need to create some information to help you identify the double crossover classes!
F1 beetle females of wild-type appearance but
heterozygous for three autosomal genes are mated with males showing three
autosomal recessive traits: oval eyes,
sleek bodies, and rippled thoraxes. The
progeny of this cross are distributed in the following phenotypic classes:
Phenotype
|
Number
|
Oval
eyes, sleek body
|
29
|
Oval
eyes, rippled thorax
|
559
|
Wild type
|
42
|
Rippled thorax
|
15
|
Sleek body
|
542
|
Oval eyes, sleek body, rippled
thorax
|
35
|
- Define appropriate gene and allele symbols according to standard conventions.
- Show the arrangement of alleles on the two homologous chromosomes in the parent (F1) females.
- Draw a genetic map based on these data.
- Can you account for double crossovers? Explain.
Another kind of trickinessness
This is the last example from the question sheet that you can download online.wo true-breeding minks were crossed. Between them there were three loci that were being analyzed: drawn jowls, club foot, and a behaviour of being easily startled. The F1 minks were all wild-type in appearance. The F1 females were testcrossed and the progeny sorted. The following data were obtained:
Phenotype
|
Number
|
drawn jowls, normal foot,
easily startled
|
157
|
normal jowls, club foot, calm
|
165
|
normal jowls, club foot,
easily startled
|
15
|
drawn jowls, club foot, calm
|
139
|
drawn jowls, club foot, easily
startled
|
14
|
normal jowls, normal foot,
calm
|
21
|
drawn jowls, normal foot, calm
|
18
|
normal jowls, normal foot,
easily startled
|
163
|
- Define appropriate gene and allele symbols according to standard conventions.
- Draw a genetic map based on these data.
- What can you infer about these loci?
(You can get a full-screen version by clicking on the "YouTube" logo at the bottom-right corner of the video above.)
Calculating Interference Two Ways
I've demonstrated above that I calculate interference by comparing the ratios of DCO offspring (observed/expected). It's simply a ratio to show what percentage of DCOs you see in the offspring compared to what you're expecting.
The McGraw-Hill Hartwell et al. textbook uses the ratio of obs DCO frequency to exp DCO frequency.
In my method, you calculate the expected DCO frequency and multiply it by the total number of offspring to figure out how many you should have seen.
In the book's method, you use the calculated DCO frequency, and convert the observed number of offspring into a frequency.
Basically, it's the same number of steps. That said, I can understand why this might be confusing and so I made a video of it (in part because the book's way of doing it confused me at the end of class and I made a fool of myself.
Here's my apology. But you can go a little further ... here are the ratios and you can calculate the distances yourself. Don't bother making new gene symbols; they did that for you. The eight rows represent all the progeny from a cross between a het and a tester.
In my video below, you see the distances, and two ways to calculate interference from it. Try it out yourself before going to the video.
Reverse Gene Mapping
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:
- 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).
- 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.
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!
Replica Plating / Interrupted Mating
In bacteria, gene mapping is accomplished differently from eukaryotes. For one thing, they're haploid, and so we're not going to see nonparentals!
Replica plating is done to look for the genotypes of exconjugants, and this presentation doesn't talk about conjugation, F-, F+, or Hfr strains. You have to know those things before this exercise. Your textbook probably doesn't go into replica plating too much, so the "Solution" to the problem below gives a bit of information on that.
Here's the exercise. Try it before looking at the solution!
And now - the solution!
Another Interrupted Mating Mapping Example
Here's another question you could practice with. Sketch a single chromosomal map with the loci spaced out properly. Include the site of Hfr insertion for each of the four strains, indicating proper polarity (direction in which it inserted).
The answer is here.
Intragenic Mapping
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!
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