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Summary of Lab 10 for BIOL 112 (Winter 2011). Week of Mar 14 2011.

Bioinformatics.

Genetics: recombination and mapping problems.

Probability-wise, you encounter ~20 amino acids before encountering a stop codon (just by chance)
Open reading frame: does not contain stop codons
- Three reading frames for each stretch of DNA for obvious reasons
- Although you could potentially have a sequence coding for three overlapping proteins (one in each ORF) it's unlikely
- If you find an ORF that codes for more than 60 amino acids then it's probably coding for a protein
Although DNA is not random, it will appear random if you try to translate it unless it codes for a protein in that reading frame
All genes start with the methionine codon (ATG), end with a stop codon (TAG etc)
We can use the SeqBuilder program to see if a given protein is similar to other proteins etc
- And even make a phylogenetic tree omg!!
Organisms have more than one DNA polymerase because there is more than one mRNA to be formed in the cell at any given time

Happens during prophase I of meiosis
When genes on the same chromosome, they are considered linked
- However, when two chromosomes (homologs) exchange material, recombination occurs
- Cis-arrangement: a+b+/ab
- Trans-arrangement: a+b/ab+
- Recombination rate depends on distance between genes - genes located close together will be recombined less often
- Recombined 1% of time --> 1 map unit away, etc
- Loci - place occupied by genes on chromosomes; correspond to map units
In Drosophila, genetic recombination occurs only in the female on autosomes and allosomes (does not occur in males)
For genetic recombination problems:
- The least common progeny are the result of recombination
- Use that to figure out what the original arrangement was (cis or trans)
- You can calculate distance (in map units) from percentages etc

Lab 10 summary