Monday, December 6, 2010

December 6th
Caroline Ryba


One more time!! Sorry...
To start class off, Mr. Paek explained what we would be doing today. He said we were going to learn about sex linkage crosses and pedigree charts.

He briefly went through a power point on pedigrees, on which we were told not to take notes. The key things we learned were that a pedigree is a type of chart, and it sort of looks like a family tree. In a pedigree chart, a male is represented by a square, and a female is represented by a circle. 
Here's an example of a pedigree chart: 



The first square that's connected to the circle, symbolizes a married couple. The two circles below them symbolize their offspring, and the square connected to the circle would be a (female) sibling. Mr Paek did not explain why some shapes are black, while others are white. 


Next, Mr. Paek remembered that in order to go further with pedigrees, we have to know about sex linkage crosses.  So, we flipped to page 45 in the UP, which focuses on sex-linkage crosses.


Two Key Things to Know About Sex Linkage Crosses:


* you will always have a mother with chromosomes XX crossed with a father with chromosomes XY. 
* the X chromosome on a male or female will always have a little letter written in the form of an exponent. the Y chromosome for a male will rarely have a little letter.
* if the problem says its on a sex linked trait, it's on the X chromosome.


Mr. Paek then told us about the three genetic disorders we should know about. They are:colorblindness, hemophilia, and Ducheme muscular dystrophy. These are caused byrecessive genes on the X chromosome. Also, women who don't have the disease, but have one recessive allele are carriers. In other words, women who are heterozygous dominant are called carriers. 


Then, Mr. Paek went through the example on page 45. 
Here's the example:


Steps 1 and 2: Assign Letters and determine the genotypes of the parents. 

Assigning letters: X^{H –} Normal, X^{h –} Hemophilia, and Y – male chromosome with no hemophilia allele.

Parental genotypes: Carrier female: X^H X^h, Normal male - X^H Y

Step 3: Determine the gamete produced by each parent

X^H X^h = X^{H ,} X^h

X^H Y= X^H, Y

Step 4: Set up Punnett square using the gamete genotypes

	XH	Xh
XH
Y

Step 5: Combine

XH Xh XH XHXH XHXh  Y XHY XhY * Sorry... i couldn't insert the table i made from word onto here, so just imagine lines in between each gene.

Steps 6 and 7: Determine the phenotypes of the offspring and state the genotypic and phenotypic ratio.

Genotype: ¼ X^H X^H , ¼ X^H X^h  , ¼ X^HY, and ¼ X^hY

Phenotype: ½ normal female, ¼ normal male, and ¼ hemophiliac male.

* So, the first row contains the two normal females. The second row, first column contains the normal male, and the second row, second column contains thehemophiliac male * 


After that, Mr. Paek did another example:

	Xh	Y
XH	XH Xh	XHY
Xh	XH Xh	XhY

The whole first row, first column box represents a girl carrier. The first row, second column box represents a normal boy. The second row, first column box represents another girl carrier, and the last box represents a boy with the disease.

Since these are recessive disorders, if the chromosome has the disorder, its represented by a lowercase letter.

* For all you who were absent, it would be really helpful to thoroughly read through page 45.*

Lastly, Mr. Paek went through a couple homework problems.
The first one was #5 on page 41 in the UP:

Question: In hamsters, yellow coat is incompletely dominant over chocolate colored coat. The heterozygous blend results in a tan colored coat. What are the genotypic and phenotypic ratios of the offspring from a cross between two tan colored hamsters?

So, we chose the letter Y to represent yellow coat, the letter C to represent chocolate coat, and YC to represent tan coat.

Here's the answer:

	Y	C
Y	YY	YC
C	YC	CC

Genotypic ratio: 1 YY: 2 YC : 1 CC
Phenotypic ratio: 1 yellow: 2 tan: 1 chocolate

Next was number 7 on that same page. Students were confused on how to determine traits for 8 offspring. Mr. Paek said to just do the problem as if you were solving for 4 offspring, then multiply your results by 2.

After that was #6 on page 44. We used the process of elimination to determine the impossible blood types for a father when the mother's and child's blood types are given.

Answer:
Mother: O, Child: B, Father can't be O or A.
Mother: B, Child: A, Father can't be O or B.
Mother: AB, Child: B, Father can have any blood type. (so no impossible blood type).

The last problem we went over was #5 on page 44. We just drew a bunch of Punnett squares to determine the right pair ups of the parents and babies.
Answer:
Parents 1 & Baby 3
Parents 2 & Baby 2
Parents 3 & Baby 1

To finish class, Mr. Paek let us work on our homework for about 10 minutes.

HOMEWORK: Pages 45 & 46 in UP.

The next scriber is...Connor because he volunteered last week.