December 7, 2010

Sribepost by: Connor “Dunbar”

HOMEWORK

-Page 49-50 in UP

-Lab 35 from lab book due Friday

Today in Class...

Today at the beginning of class Mr. Paek briefly went over sex traits for the many people who were out of class the day before. To see the notes on sex traits view the previous blog post by Caroline R. Today we went through Mr. Paeks powerpoint on Pedigrees. A pedigree is like a family tree without names, but you can trace disorders or traits throughout a family using one. Scientists use these charts to study patterns of inheritance within a family. Some general information you should know about pedigrees is a male is shown by a square and a female by a circle. Children are shown by a line drawn down from the parents that branches and connects the children to the parents. Below is a picture of an autosomal pedigree chart.

We then learned how to interpret a pedigree chart. First we learned how to interpret disorders shown in a chart. A shaded person has the disorder. If mostly males are infected with the disorder, it is usually a X linked trait or sex linked which means this disorder is found on the 23 chromosome. If the disorder is 50-50 between men and women, the disorder is autosomal. This disorder or trait is then found on chromosomes 1-22.

This is a X or Sex linked trait

Next we learned how to determine if a trait was either dominant of recessive. If the trait is dominant, one of the parents must have the disorder for the next generation to get the disorder. If the trait is recessive parents do not need to have the disorder because it could be heterozygous making them a carrier but not showing the disorder. If this is the case the disorder may skip generations. We then did a few examples in our unit packets pages 47 and 48. I briefly will describe the first pedigree chart on page 47.

1. Person A is a female because her sign on the chart was a circle. Her phenotype was attached because her circle was shaded in meaning she had the recessive gene, in this case attached ear lobs.

1. We had to find the phenotype of person B, which was Ee. We determined this by looking at his children. Two children were not shaded meaning they had unattached earlobes and one was shaded meaning he had attached earlobes. His wife had ee so he had to have Ee so the recessive genes could pair up to give the child attached earlobes. It would be impossible for that child to have attached earlobes if the fathers genes were EE

1. For number three it asked what the genotype and phenotype of person D was. THe Phenotype was free hanging because when looking at her parents she would have gotten the dominant from her fathers Ee and the recessive from her mother. So, her genotype was Ee giving her unattached earlobes.

1. The final question was asking if 2 people with free hanging earlobes were to have a child would it be possible for the child to have attached earlobes. It would be because if both parents were heterozygous the recessive genes would meet in the last square of the Punnett square. This would give a 25% chance for one of the children to have attached earlobes.

This was about it for today. The next scriber will be...........DAMBI. Sorry.