Werecreatures! Scaaaary!

I took some questions designed by Dr. Gillian Davis and fleshed them out a bit.  Here they are.










For each

• Define appropriate gene and allele symbols according to Advanced Nomenclature rules.
•  Fill in the table. The genotypes must reflect the correct gene order (and b/c they are gametes, should be haploid genotypes from the het parent!!!)
• Calculate the map distances between the loci (correcting for double crossovers) and draw the map.

See the solutions below - the Werebat is detailed, and I breeze through the other beasties quickly in the second video.

...And the others ...

"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!

Determining Gene Order

I had a lot of office visits today regarding 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+

You should note that in both cases, the genotypes of the F₁ are the same:  they both represent heterozygous creatures.  This will dramatically change the ratios from your testcross and which numbers represent the "parentals" (which actually just give the chromosomes for your F₁).

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 parent

For 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).

Chromosomal Rearrangements

Here's a worksheet for you to practice with.

A solution video is below.  I had trouble uploading it, so it's in lower-quality and I'm not sure it started at the exact right point.  It should work for you, though.  You can make it larger by clicking on the YouTube logo in the bottom-right corner of the embedded file and running it straight from YouTube.

Excellent videos explaining the Meselson-Radding model of recombination

The molecular basis of recombination is amazingly precise:  crossover is so exact that not one atom of homologous chromosomes is added or deleted between the two double-helices that are being recombined.  The mechanism is directed between base-pairing between the nonsister homologous chromosomes.

These videos demonstrate how the molecular mechanism of recombination can also be responsible for gene conversion.  The organism in which gene conversion has been best studied is a mold that keeps its spores in a sac called an "ascus".  As meiosis progresses, the cells are kept in a linear array which allows the scientist to follow the fate of each cell.  You can map a gene with respect to its position from a centromere using this system.

But gene conversion, not mapping, is what I want to describe.  When meiosis occurs, you should have equal numbers of each kind of allele when you end.  For example, if you have a B and a b allele of a gene, the end product of meiosis, four cells, will contain equal numbers of B and b two cells will be B and two will be b.  Recombination will assort the alleles, but you're not gaining or losing any genetic material.  If the haploid cells at the end of meiosis undergo mitosis, you'll double the number of both alleles:  you'll get four cells that contain B and four that contain b.

Robin Holliday, a geneticist, was studying asci and mapping out how the alleles segregated.  He noticed a strange phenomenon, though... sometimes instead of getting 4 B alleles and 4 b alleles, he got 5 B and 3 b.  Sometimes it was 6 B and 2 b.  Even the reverse happened: 3 B and 5 b.  Sometimes it was 2 B and 6 b.  Apparently one of the alleles got changed - converted - to the other allele in the cross.

Holliday recognized that the presence of both alleles together in the zygote might lead to them interacting.  Crossover is when nonsister chromatids are more likely to interact, and he thought of a model where the double-helix of each chromatid might unzip and then base-pair with its nonsister partner.  They'd have essentially the same nucleotide order, except for the region that differs to make them allelic and not identical, and so base-pairing can connect these nonsisters together.  They'll be cut apart to separate, and depending how you cut them, you might cause a crossover event in which new allele combinations on a chromosome result.  

Here's a video (courtesy of Brooker's Genetics, 2nd edition, McGraw-Hill)
... also ...

Holliday's model has inherent symmetry, though, and the converted asci don't typically have symmetry in their numbers.  5:3 is not symmetrical.  Matthew Meselson and his colleague Charles Radding modified Holliday's theory by introducing assymetry:  only one strand is cut and it dislodges the partner strand of the nonsister chromatid.

A further model, double-strand break repair, is a further refinement of the crossover model and is considered to be the most likely mechanism, although from what I hear, there are differences between species.

I like this concept as it demonstrates the progression of science, and how we can use concepts from different parts of the discipline to inform the other parts.  This is a great combination of how classical and molecular genetics can cross-pollinate!

Question 1a

Well, on to the first lesson!

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.

---

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

 

1. Define appropriate gene and allele symbols according to standard conventions.
2.  Map the distance between the two genes.

---

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).

Question 1b

The first example was pretty straightforward:  two genes.

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

1. Diagram the arrangement of alleles on the two homologous chromosomes for both parents (P generation) and the F₁.
2. Draw a genetic map based on these data.  Be sure to mathematically correct for double-crossovers.
3. 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).

Question 2

This 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

---

1. Define appropriate gene and allele symbols according to standard conventions.
2. Diagram the arrangement of alleles on the two homologous chromosomes for both parents (P generation) and the F₁.
3. Draw a genetic map based on these data.  Be sure to mathematically correct for double-crossovers.

Try this question before looking at the video solution below:

(you can get a full-screen version by clicking on the YouTube logo at the bottom-right hand side of the video above).

Question 3

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

1. Define appropriate gene and allele symbols according to standard conventions.
2.  Show the arrangement of alleles on the two homologous chromosomes in the parent (F₁) females.
3. Draw a genetic map based on these data. 
4.  Can you account for double crossovers?  Explain.

---

The video solution is below:

(you can get a full-screen version by clicking on the YouTube logo at the bottom-right hand side of the video above).

Question 4

This is the last example from the question sheet that you can download online.

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Two 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

1. Define appropriate gene and allele symbols according to standard conventions.
2. Draw a genetic map based on these data.
3. What can you infer about these loci?

---

My students often find that this one is tricky!  See if you can spot the hard part before you view the solution below.

(You can get a full-screen version by clicking on the "YouTube" logo at the bottom-right corner of the video above.)

Alien gene mapping

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.}