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