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.
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.
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 taughtas 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.
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.
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!
Names are handles to help us organize information. We can use given names or nick names to identify particular people. We use species names or common names to identify particular types of organisms. Names help us find locations: house numbers and streets, intersections, landmarks. Sometimes names give us insight into details about a specific edifice. For instance, if I'm told of a building and its name is St. Andrews, I have a good idea that it's a church or part of the Separate School system in my city. The words "University", "College", or "High School" tell us that we're looking for an educational venue.
Geneticists don’t work in isolation – they are part of a wider community of fellow geneticists. Many geneticists tend to communicate most often with other researchers working on their same organism. To make things easier, they agree on how to name mutant loci so that it's faster and more clear when they communicate their findings. Choosing a good naming convention can save an immense amount of time. You'll see that the naming convention described here allows you to just look at a gene symbol and you can tell if it's X-linked, autosomal, dominant, or recessive. If you see two gene symbols, you can tell if they're on the same chromosome (linked) or on separate ones. It's a powerful, efficient system.
Pretend we are a community who will all explore a hypothetical organism: a groo. To do this we will use a common naming and symbolic system. We’ll call it the “Groo Convention”. Groos are dimorphic (males look different from females), are brick-red in colour, and have a blue, furry surface. Here's what wild type groos look like (guess which is the male and which is the female!):
Because the groos above are well-described by their researchers, the normal phenotypes (those most often found in the wild, and hence "wild type") are taken for granted. Deviations from these - mutations - are therefore interesting and are likely to be studied to determine the genetic nature of these. Thus our first rule:
1) The "normal" groo phenotype is described as wildtype, and this is the "base state" or "reference organism".
2) Any change from the wildtype phenotype is a mutation and must be named. For example, a trait that makes a groo appear as very pale could be named “pale groos.” Note that names should be descriptive: cute puns/jokes might work (maybe you want to call the mutants below jaundice), but the name provides a clear description of the mutant trait.
3) The gene symbol must be 3 letters derived from the trait name. Underline the letters to show that they belong to a single trait. For example, you might want a "p", an "a", and a "g" from pale groos.
4) Dominant mutant trait symbols must have the first letter capitalized. Recessive mutant traits must be all lower case. Mutants are not always recessive!! Sometimes they are dominant! For example, if one determines that the pale groos trait is due to a dominant mutation, we write:
Pag
Or, if the pale groos trait is found to be recessive, we would write:
pag
5) The allele that is not mutated in this locus uses the exact same letters as the mutant, only a superscript ‘+” is added to the symbol. For example, assuming the pale groos mutation is dominant, the wildtype allele of Pag is:
Pag+
6) Homologous chromosome pairs are indicated with a ‘/’ (slash) symbol. For example, a groo that was heterozygous for the pale groos mutation, write:
Pag / Pag+
7) Genes on nonhomologous chromosomes are separated by a ‘;’ (semicolon) symbol. For example, a groo that was homozygous for the pale groos mutation and heterozygous for a different trait, blue, on a different chromosome, write:
Pag / Pag ; blu / blu+
*this image was from another resource where I named a locus "Paf". Please note it should be "Pag" for this example.
8) Linked loci (loci that are on the same chromosome) are listed together on the same homologous chromosome. For example, if pale groos and being bald were linked traits, they would be on the same chromosome, and thus a groo that was homozygous mutant for Pag but heterozygous for blu would be described as:
Pagblu+ / Pagblu
*this image was from another resource where I named a locus "Paf". Please note it should be "Pag" for this example.
9) For X-linked traits (sex-linked), the trait is written as a superscript to the letter X. For example, if vermillion was a dominant sex-linked trait, write:
Xver (mutant) or Xver+ (wildtype allele of vermillion)
*note: the underline for "ver" should be IN the superscript. The web editor does not force this for every browser. Here's what it should look like:
10) If the groo is hemizygous (i.e. male), use a ‘Y’ in place of the homologous chromosome. For example, if there was a male groo with the X-linked trait, vermillion, write:
Xver /Y
11) If the groo has two X-linked genes, add the loci to the superscript only. For example, if there was a female groo heterozygous for two recessive X-linked traits, there are two possible genotypes:
Xver+pub+/ Xverpub Xver+pub / Xverpub+
… and let’s look at a female who is heterozygous for a dominant mutant allele and a recessive mutant allele. The genotypes could be:
XGlu+pub / XGlupub+or
XGlu+pub+ / XGlupub
What phenotypes would you write down for these? The first would be a wild type female (het for two recessive mutants) and the second would appear to be a female Glu groo.
Let's try this out with some examples.
From the genotype shown, determine
i. How many mutant traits are being described in this groo?
ii. The inheritance pattern of each trait (sex-linked, dominant or recessive),
iii. The linkage, if any, of traits,
iv. The phenotype (including sex) of the groo, and
v. The zygosity of each trait (homozygous [dom/rec], heterozygous, hemizygous).
The genotype:
Xhel / Y ; grn/ grn+ ; Brs / Brs+
Now let's go the other way. Give all possible genotypes for a grooinwhich:
i. Three traits are involved, tny, wht, and Eye
ii. Eye is X-linked; tny, wht are autosomal.
iii. This groo is heterozygous for all loci.
Bonus question:What is the phenotype of this groo?
