February 2018

# Genetics review.

This is info on the upcoming test, 2/26. Read More…

# Exercises for lab period

We will be doing an exercise on chi-squared which you can find here. Also, the AP Bio equation sheet is here.
My answers to that mosquito question is found here.

Then, we will do an exercise on Population Genetics and the Hardy-Weinberg equation. You can find that here. Read More…

# Karyotype and Pedigree

This entry covers karyotype and pedigree. We will do the online exercises tomorrow in class. Read More…

# Link to stuff on the smart board today.

I captured the stuff from the white board today and made it into a web-page. Read More…

# Chi-squared followup

## Chi-squared followup.

Because I expected to get farther before the test, I have blog posts that go into the things I meant to cover. So, I'll move them closer to the top so you can see them (the will be just below this one).
I made a dumb mistake as you were walking about the door.
I gave you the wrong definition of "pleiotropy." (what I gave you was the definition of "epistasis," another new term. I have them properly described/defined, in the blogs below. Please make note of my mistake if you wrote it into your notes.

Below are some data to practice some more chi squared calculations, if you would like to do them.

This is a Dihybrid cross, meaning AaBb x AaBa. That means the expected results are 9:3:3:1. That won’t be what you get.
Cage 17 is the F1 dihybrid. Cages 12 and 16 are their because the were the source of the P1 generation.The remaining three cages, 18, 21 and 25 are just increasing numbers of offspring. Try performing a chi-square for 18, 21 and 25. It’s pretty striking how the size of the data set makes such a difference. There is a chi-square table below to assess the probabilities. For the homework, submit the chi-square value for each cage. What are the “Parental” types you expect to see more frequently if the genes are linked. Do you think the genes are linked? Explain your reasoning.

# Plieotropy, multi-gene traits and epistasis

## 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:

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
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 determines 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.

## Single Locus

As we've learned, if there are two alleles at one locus not on sex chromosome, the hybrid cross (Aa x Aa) must produce the 1:2:1 assortment. However, the 3:1 dominance is only predicted if there is a simple dominance relationship. aa x aa is boring. The other three types of crosses are shown below, AA x aa; Aa x aa and Aa x Aa

Here's how that maps out for a simple dominance on the X chromosome:

For incomplete dominance, the case is interesting because the heterozygous phenotype ONLY shows up in females.

## Multiple Loci

The first think is how do we deal with multiple loci? There is a way that is done purely with math…that's what we do when there are more than two loci. However, I'm going to show you how to do it with a punnet square.