Wednesday 30 November 2011

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!

Tuesday 8 November 2011

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.


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.