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More Karyotype etc

Here is a link to a site at Stanford that helps you understand pedigree. If the link downloads a file instead of opening a page, you have to do the following: right-click or control-click the link and select "copy link." Then paste it into an address line and change the ending from "swf" to "html." Then it should work.
There is a short quiz you can take. I won’t see the results.
and here is the Karyotype activity
Below is a blog I wrote on the topics. You can use it as a reference if it helps.

Karyotyping

:
(as usual, images from wikipedia).
Karyotyping is a technique for looking at all the chromosomes in a cell. Briefly, a pro-metaphase cell is is stained with a dye that binds chromosomes. There the traditional dye, giemsa, stains the chromosomes in a banding pattern that is different for each one. So, the banding pattern of chromosome 1 is different from that of chromosome 2, etc. These days, there are specific color-coded probes that make it easier. The cell is then photographed and the picture is enlarged. Then, traditionally by literally cutting up the picture, but now digitally, the chromosomes are cut out and aligned so that we can see that the individual has 2 copies (one from mom and one from dad) of each of the 22
autosomes. The largest is chromosome 1 and the smallest is chromosome 22.
Karyotype
Then look at the last pair: The really tiny one, smaller even than chromosome 22, is the “Y” chromosome (so this is a male). The larger one next to it is the X.
Look how much cooler it is with the modern technique:
ColorKaryotype

As you can see, this individual is a female.
Using techniques such as this, we can look for large changes to the DNA, rearrangements of the chromosomes we call “inversions,” “deletions” (both self explanatory), translocations (where a piece of one chromosome is spliced to another), extra or missing copies of one chromosome (“trisomy” or “monosomy”) or, even whole extra sets of chromosomes.
Some common disease states result from this. Trisomy 21 causes Downs syndrome.
This can occur when the sister chromosomes of 21 do not separate at meiosis I (this is called “nondisjunction”). That results in a 2 gametes getting no chromosome 21 and 2 others getting 2 copies of it. If one of that latter group fuses with a normal gamete, the resulting zygote has 3 copies.
Check out this karyotype:
XXY
Do you see what’s wrong? This person has Kleinfelter’s syndrome, where he has 2 X chromosome and a Y.

Pedigree Analysis:


Here is the pedigree example given in the Campbell Book.
Widows
Females are circles, males are squares. People with the trait being followed usually are filled in with a color, people without the trait are not. This trait, the appearance of the “widow’s peak,” is a dominant allele on one of the Autosomes (not sex linked). How can I tell? Remember that the recessive allele can be covered up by a dominant allele. So, two people with the trait, if it is dominant, can produce a child without it. Notice that this happens in the mating in the second generation. Two “widow’s peak” people had a non widow’s peak child. You never see enough children to invoke statistical predictions in a generation. You should see only ¼ homozygous recessive in the 3rd generation…but there are only two kids. Even if there were 4, the numbers are not good enough to make predictions.
I know it is not on the X because an affected mother in generation 1 married an unaffected man and had an affected daughter. That would be impossible for a recessive allele on the X (the daughter would have to have 2 copies, one from each parent. But Dad cannot have it since he does not show the trait). I know it is not on the Y because it appears in women.

Here is the pedigree for hemophilia (a severe bleeding disease) in the royal families of Europe. The mutation seems to have occurred in Queen Victoria.
  1. Is this trait sex-linked?
  2. How do you know?
  3. How do we know that Irene (number “3”) is a carrier?
  4. Is it possible that Queen Elizabeth is a carrier?
  5. Could the disease show up in any of the children Prince William and Kate Middleton might have?
HemopheliaPedigree

Evolution Project

There is a file in the downloads folder by Neil Shubin and Sean Barry Carroll that I would like you to read most of it over the course of the project. There are lots of pointers to help out whomever does eyes and limbs. Start by reading the introduction and then we will use the rest as a sort of reference as we go.
This blog contains some suggestions for projects.
Read More...

Classes of Punnett Squares

Here you will find the examples of the "classes" of Punnett squares. That is, how the alleles will assort depending on the parents genotypes.
Also, here is the handout for today.

Here's an easy read about some recent work on a classic "ring species," where populations have diverged and speciation has occurred from one original population.

More genetics terms

Some variants in inheritance.


Multi-gene trait. There's really not much to this. Some traits…really most interesting traits…may be inherited, but be based on more than one gene. So, there is no single gene for being tall. There is no allele for being 6' tall. But, that doesn't mean that height is not inherited. Many different loci may contribute to it. Here is a hypothetical plot of what you might expect if there are three loci that contribute to height, each with two alleles where the dominant of each allele contributes positively to height (note, there is no reason any of those assumptions should be true). You might see a distribution that looks something like this:
Multigenedist

So, you can get a distribution of heights based on some combination of alleles at different loci. In this case, we are assuming simple additive interactions among the genes. It an be more complex (See below(.

Pleiotropy (note, even after looking it up, I spelled it wrong)
This is sort of the opposite of multi-gene trait. Here, one gene can affect many traits. We've discussed this in the context of cytoskeletal proteins before. For example, mutations to a microtubule-associated motor protein could affect male fertility (sperm flagellum), airway function (cilia on airway epithelium) and vesicle transport and secretion of proteins. Everywhere that protein is needed, you would see some effect.

Epistasis
Finally, there is epistasis. Proteins interact with other proteins, so variations in one gene can affect how you see the phenotype caused by another. Here is a classic example I stole from another website at the university of Georgia. It covers coat color in.
Labrador retrievers. One locus, the B locus, controls the color of the pigment eumelanin. Eumelanin can be either brown in color (bb) or black BB or Bb.
Another gene, known as the "E" locus (for extensor…never mind) is needed to deposit eumelanin in the fur of the dog. It encodes a protein called MC1R and where it is expressed determine whether the eumelanin gets into the fur. The ee homozygote does not deposit eumelanin at all in the fur while Ee or EE do.
Thus, if the dog is ee, it will be yellow no matter whether it makes black or brown melanin. All the possibilities on the 4x4 matrix are shown below.

Epistasis



Some more problems

There are several problems we can work on together. This first one is pretty straightforward.

Here is the
first one I want you to use.

The second one.

Shared problem

Here is a link to the file you should download. I also have it in a google drive folder. If the first link doesn't work, try this one.

You cannot open it by double clicking. You have to put it somewhere you can find it and then open it from within VGL2.
It should open with cage 1 and cage 8 visible. If not, go to "utilities" and select "show/hide cages" and select cages 1 and 8.
The cross in cage 8 is very telling. You should be able to tell: which of the alleles at each locus is dominant and what the genotypes are at both loci for both parents of cage 8. Using Aa and Bb assigned appropriately, determine what those are.

Next, perform a dihybrid cross (hint: cage 8 contains all your dihybrid Aa Bb)
Based on the results, can you say that the two loci are linked? Can you say they are NOT linked? Perform a chi-squared test to test your hypothesis.
Here is the
APBio equation sheet.

Next, perform a test cross (a dihybrid from cage 8 by the homozygous recessive…you need to know which that is).
Evaluate again whether you think linkage is occurring (another chi-square). Hmmm….

Here is the link to the
google form for your answers.

Important admission of error follows


Oh…this is important:
I screwed up. For standard analysis on bulk recombination, it is the percentage of recombinant progeny that you use to get recombination frequency. So, the recombination frequency should have been 28, not 14 percent. The reason for my mistake is honest, if you care. I learned genetics on an organism called chlamydomonas, where you actually grow up all four products of each meiosis as haploid organisms. So, each recombinant meiotic event (or recombinant "tetrad") produces two recombinant offspring (Ab and aB).
While that's true in a bulk cross as well, you don't assume you are seeing both recombinant progeny from each cross over. The statistics are treated differently. But, if you had a test cross with 100 individuals and the genes are far apart on the same chromosome such that they effectively randomly assort, you would expect 25:25:25:25 or "50% recombination."
It's just much cleaner in chlamy.

Intro To Genetics

This has a lot of the terms we will need and some basics of chromosomal inheritance. Read More...

Dihybrid and test cross

Dihybrid Cross.


Consider two loci, A and B, each with two alleles, A and a or B and b.
By definition, a dihybrid is heterozygous at each of two loci (AaBb).
A critical first step is to get "true breeding" flies of each trait. So, you want a set of flies that are homozygous recessive (always easy to find, once you know what the recessive is) and homozygous for the dominant, which requires test cross by a recessive to make sure. So, I want AA, BB x aabb OR AA,bb x aa,BB (in both cases, both parents are homozygous for both loci.

Below is a Punnett square set up for a dihybrid crossed by another dihybrid. The outcome of that cross will be 9:3:3:1 for phenotype as indicated.
Screen Shot 2015-01-26 at 3.01.20 PM

Here is the far simpler cross of a dihybrid by a double homozygous recessive. When you use a homozygous recessive, it is known as a "Test cross," as in the ones we did to determine if a fly with the dominant phenotype was heterozygous or homozygous.
Screen Shot 2015-01-26 at 3.01.31 PM

I did a cross in VGL of flies that are either forked or knobbed wing and black or red eye. The results of some crosses are below.
Screen Shot 2015-01-26 at 3.11.04 PM

What does Cage four tell you? Which alleles are dominant and which alleles are recessive?
If the progeny in cage 4 are the dihybrids (they are) what are the "parentals?" (just look at the parents of that cage…)
Here's the key question: are the results of the test cross (cage 8) what you expect? If not, what could account for that.

RNA Processing

A look at what happens after transcription. Read More...

RNA Splicing

I’m going to rely on a video and some details from today.
You should also be reading chapter 18 in the book. Read More...

Transcription 1

Just the beginning of the discussion. Read More...