Previous Knowledge¶

Computer Knowledge¶

No previous computer knowledge is required to follow this course, beside basic computer usage. But we expect the student to have some general biological knowledge.

Basic statistics¶

Biology Fundamentals¶

The student is expected to know the nomenclature used to classify the living world, especially the basic binomial nomenclature used in taxonomy.

Mendelian Genetics¶

The classical and molecular concept of the gene. The mendelian laws and the common exceptions to the laws.

Molecular Biology¶

What are the DNA, RNA and the proteins?
What is a genome?

DNA strands¶

The DNA is a polymer comprised by nucleotides, the DNA sequence is the sequence of nucleotides. The DNA molecule has a direction defined by the 5’ and 3’ deoxyribose sugar carbons. The 5’ carbon is bonded to the phosphate group and the 3’ carbon is bonded to the hydroxy group. The nucleotide sequences 5’-ACTG-3’ and 3’-ACTG-5’ are different. To solve this problem, by convention, the nucleotide sequences are always written in the 5’-3’ direction.

The DNA is usually double stranded and the nucleotides Adenine and Timine and Guanine and Citosine are complementary. So a complete double stranded DNA molecule could be written as:

5' - ACTCTACA - 3'
3' - TGAGATGT - 5'

But, that sequence would be written by convention just as: ACTCTACA and the sequence TGAGATGT would be a different one. The reverse and complementary sequence of ACTCTACA (that would be the other strand of the same DNA sequence in its 5’ to 3’ direction) would be: TGTAGAGT.

Transcription¶

The transcription is the process by which the genomic DNA is copied to messenger RNA. Only one strand is transcribed and the transcription starts at the promoter and goes in the 5’ to 3’ DNA direction.

The immature eukariotic messenger RNA is spliced removing the introns before becoming mature RNA.

The eukariotic gene sequence can be divided in regions: promoter, 5’ UTR, exons, introns, 3’ UTR and polyadelination site. The poly-A sequence is not present in the genome, it is added during the maturation of the messenger RNA.

Translation¶

In the translation the ribosomes produce the proteins by using the genetic code. The genetic code is almost universal, there are just minimal variations between very phylogenetically distant organisms. The genetic code is denegerated, several codons code for the same aminoacid.

The translation starts at the the start codon, that has to be an ATG codon, and proceeds in the 5’ to 3’ RNA direction. The translation ends when reaches any stop codon.

Every DNA sequence has 6 possible reading frames

forward

5' ACAAGATGCCATTGTCCCCCGGCC 3'

1  ACA AGA TGC CAT TGT CCC CCG GCC

2  A CAA GAT GCC ATT GTC CCC CGG CC

3  AC AAG ATG CCA TTG TCC CCC GGC C

reverse

5' GGCCGGGGGACAATGGCATCTTGT 3'

4  GGC CGG GGG ACA ATG GCA TCT TGT

5  G GCC GGG GGA CAA TGG CAT CTT GT

6  GG CCG GGG GAC AAT GGC ATC TTG T

An Open Reading Frame (ORF) goes from the start codon to the stop codon.

Proteins¶

A protein is a folded peptide. A peptide is an aminoacid chain. The aminoacid chain has a direction, it starts at N terminal and ends at C terminal.

Molecular genetics technologies¶

Basic Genetic engineering technologies: restrictions enzymes, ligases, DNA cloning, cloning vectors and PCR.

Sequencing technologies.

Bibliography¶

Concepts of genetics. Klug, William S.; Cummings, Michael R.; Spencer, Charlotte A. Pearson, cop. 2014. 10th ed.

Campbell biology. Reece, Jane B; Urry, Lisa A.; Cain, Michael L.; Wasserman, Steven A.; Minorsky, Peter V.; Jackson, Robert B . Pearson, cop. 2014. 10th ed.

Molecular biology of the gene. Watson, James D. Pearson, 2004. 5th. ed.