Scribe Post 12/8/10

Announcements: Extra Credit 64 Punnett square due Monday; Quiz on Monday



Homework: Thursday: UP pages 51-52

Friday: Lab 35

Monday: UP pages 53-56



What We Did In Class: We talked about the different syndromes that can be caused with extra chromosomes and we went over UP pages 51-52.



Syndromes:



Edwards Syndrome: an extra part to the 18th chromosome. It has many physical and mental defects that it can lead to and early death.



Patau Syndrome: an extra part to the 13th chromosome. It has the same defects as Edwards Syndrome, only more severe, that it can lead to death soon after a baby is born.



Turner's Syndrome: a person that is missing a 2nd sex chromosome (ex. XX, XY now becomes X, X). It has minor conditions, but they cannot produce offspring.



Down Syndrome: this person has 47 chromosomes instead of 46. About half of the people born with Down Syndrome have heart problems at birth, so some can lead to early death.



Klinefelter's Syndrome: this person has an extra sex chromosome (ex. XXX, XXY). They are usually male. They have a low level of testosterone and usually have female-like features, since they also have 2 X chromosomes.







UP pages 51-52:



We talked about how children can obtain diseases from their parents in their genes. For the Sickle Cell Disease we found out that the alleles were codominant, which basically means dominant. For the Cystic Fibrosis, we found out that it's caused by a recessive allele in the 7th chromosome. For PKU and Tay-Sachs Disease, we found out that they both are autosomal recessive, so if they had children with people without the syndrome, then their children will most likely not have it either. However, for Huntington's disease, is autosomal dominant, so even if they had children with people that don't have the disease, they will still most likely get the disease, too. For Achondroplasia, it is also dominant, so their kids would most likely have the trait.

Well, that's all we did today, I hope everyone enjoyed it and got as much as they expected out of me.

The next scriber is Drake. Good Luck!!!

Scribe Post 12.7.2010

12.7.2010

Announcements: Quiz Wednesday or Thursday.

Homework: Finish Lab 35 (Due Friday) & 49-50 if not yet done.

Today’s period consisted of going over the homework Mr. Paek had assigned yesterday (12.6.2010) and starting Lab 35 in class.

• For example, one problem that we went over was (pg. 48 practice)...



Question: What is the probable genotype of individual D?

Solution: Since we know D’s parents are Hh x hh we can now make a Punnett square.



To determine which one genotype is correct, we must look at D’s offspring. Individual D’s husband has a genotype of hh. One of their children, K, has Hh. The Punnett squares below show two possible crosses for C x D given that D is either Hh or hh.





Therefore, since it is not possible to have an offspring who is Hh (K), the genotype for D must be Hh.

Also, we went over what we had learned so far such as the single traits, dihybrid, codominant, blood type, and sex linked genes. (Briefly explained below)


• Single trait- a trait controlled by a single gene that has two alleles.



• Dihybrid- mating between two individuals, both of whom are heterozygous for the two genes being followed. i.e: AaBb x AaBb




• Codominant- Describes two or more alleles that are equally dominant.
i.e: WR- WR= White + Red



• Blood Type- Contains three different alleles, two copies of which exist in all human body cells.




• Sex Linked Genes- concerning characteristics that are determined by genes carried on the sex chromosomes. (X & Y)



This concludes our Bio day :)


Next Scriber- Lauren

Monday 12.6.10

Announcements: None

Homework: UP pages 46-50

Today in class we learned about X-Linkage (Sex Linkage) Problems. An X-linked trait is one whose gene is carried only on the X chromosome. Females have two X chromosomes (XX) and males have one (XY). There is no corresponding gene on the X chromosome, so one recessive trait could show up in a male. Examples are colorblindness, hemophilia, and Duchenne muscular dystrophy. Women who don't have the disease but have one recessive allele are called carriers.

Affected female has children with a normal male.

Steps 1&2: The first thing you do is assign letters and determine the genotypes of the parents. XM-normal, Xm=affected, and Y=male chromosome with no affected ellele.
            XMXm-carrier female, XMY-normal male

Step 3: Determine gamete genotypes prodeuced by each parent:
            XMXm= XM, Xm           XMY= XM, y     
For all X-linked punnett squares, always start with a normal punnett square and XY (male) on one side and XX (female) on the other:


Step 4: Set up the punnett square with XM and Y on one side, and Xm and XM on the other

Step 5: Combine parental gametes:


Genotype: XMY*XMXm
Phenotype: 1 normal female (XMXM), 1 normal male (XMY), 1 carrier female (XmXM), and 1 affected male (XmY).

Autosomal: any chromosome other than a sex chromosome.


We also learned about Pedigrees.
SYMBOLS:
Squares=male
Circles=female
Black square/circle=affected (recessive) individual
Half black/half white=carrier


Here is an example:

Pedigrees can be x-linked or not x-linked. In the pedigree above, the black circles/sqaures represent colorblindnes, it is recessive. Colorblindness is x-linked. For the pair in the first generation, the male is colorblind (XcY) and the female is not. She is not a carrier, so she must be XCXC. Their sons are not colorblind, so they are both XCY. The daughter in the second generation is a carrier, so her genotype is XcXC. Her husband is a normal male (XCY), and so is one of her sons. The other son got the recessive gene so he is colorblind (XcY).

Here is a pedigree for a trait that is not x-linked:



     This is a pedigree for attached earlobes (aa). For the first generation pair on the left, the male has attached earlobes (aa) and the female does not. The female must have the recessive gene (Aa) because one of her children has attached earlobes. The one that has free hanging earlobes also must have the recessive gene (Aa) because it has a kid with attached earlobes.
     The first generation couple on the right both have the genotypes Aa. This is because 3 out of their 4 kids have attached earlobes. The one that has free-hanging earlobes has Aa, because it has a kid with attached earlobes.


Next scriber-Gabriela

Friday 12/03/10

Anouncements: Be ready for a Pop Quiz. Next week for Extra Credit Mr.Paek will give a trihybrid cross. (64 squares)

Homework: UP 40, 41, 44

What we did in class...
We did UP pg. 39 and pg. 43. We learned how to do Incomplete Dominance, Codominance, and Multiple Alleles.

Incomplete Dominance:
*Sample Problem: Cross a red flower with a pink flower.

• Assign letters (R- red W- white RW- pink)
• Determine the parental genotypes. (RR x RW)
• Set up a Punnet Square using the gamete genotypes.
  
  
  

• Combine the gamete genotypes of one parent with those of the other parent to show all possible offspring genotypes.
  
• 
  
• 
  

• Determine the phenotypes of the offspring and state the genotypic and phenotypic ratios.
  --- Genotype: 2 RR, 2 RW
  --- Phenotype: 2 Red, 2 Pink

Multiple Alleles:

*Sample Problem: Cross a homozygous for type B blood with a heterozygous for type A blood.

• Assign letters and determine the genotypes of the parents (IBIB x IAi)
• Determine the gamete genotypes produced by each parent (IBIB= IB, IB/IAi= IA, i)
• Set up Punnett square using the gamete genotypes
  

• Combine parental gametes
• 
  

• 
  
• 
  

• Determine the phenotypic and genotypic ratios of the offspring and answer the question. --- Genotype: 2 IAIB, 2 IBi
  --- Phenotype: 2 type AB blood, 2 type B (hybrid) blood

Joels Scribe for 12/2/2010

Hey everyone, here's what we did today.

HOMEWORK!
35+37-38

Announcements:none

Class Time:
In class we did not do much, we talked about Dihydrid problems and then Mr.Paek gave us time to work on the unit packets.
-We were talking about crossing over, that's when two chromosomes pair up, get very close and exchange parts of each. There are 5 different kinds, we learned the second kind.




-When you try to find out the hair color or height, more then one set then you use Dihydrid. Most commonly, when you have two traits.






How to Solve Dihydrid problems


(1) Find the Genotypes and Phenotypes


Ex:Tall=T, Short=t





(2)Determine the parent genotypes


Ex:TTtt or RRrr



(3) Use the F O I L method like Mr. Paek showed us


FOIL=First,Outer, Inner,Last





(4)Put in on a Chart Like this....





















(5)Then find out the Genotype and Phenotype Ratio

That's it!

Tomorrows Scribe will be by....Zack(if he did it, then Alex)
I hope i was helpful, goodnight to all!

JAKE NELSON, 12/1/10

Announcements: none
Homework: finish pages 23-28 if you did not finish them in class

Mr.Paek briefly went over meiosis with us for a couple minutes and showed us a slide of the stages within meiosis.
Meiosis 1
Prophase 1- KNOW THIS IS THE STAGE WHERE CROSS OVER HAPPENS
Crossing over is when pieces of homologous chromosomes cross over and switch place, this results in variety of the offspring.
Metaphase 1
Anaphase 1
Telephase 1
Meiosis 2
Prophase 2
Metaphase 2
Anaphase 2
Telephase 2


Then Mr.Paek told us to do the lab on pages 23-28, which had to do with genotypes and traits. Heads means that it is a dominant trait (A). Tails means that it is a recessive, buy the only way it can be a recessive trait is if both the coins land on tails. That was our day in biology class.