Chapter 8: Microbial Genetics

Question 1

DNA

Answer 1

1. made of nucleotides
2. double helical structure held together by hydrogen bonds (A=T) (C triple bond G).
   - polynucleotides are complementary in nature
3. what is a gene?
   - a segment of DNA
   - it codes for a functional product (RNA protein)
   - product is usually a mRNA
   - alternatively, the product could be a rRNA, or tRNA
4. antiparallel natures
   - one strand of DNA structure is upside down relative to the other
   - if the sugar is on top in the first strand, it will be on the bottom of the second
   - 5-3 is the leading strand
   - 3-5 is the lagging strand
5. Semi conservative replication
   - anabolic rxn
   a. enzyme DNA helicase unzips
   b. enzyme DNA gyrase relieves the tension of trying to coil the genes
   c. open section forms a replication fork
   d. leading strand
   - enzyme DNA polymerase (large) build a new strand of DNA
   e. lagging strand
   i. Enzyme RNA polymerase (small) forms a RNA primer
   ii. enzyme DNA polymerase removes the RNA primer
   - (exonuclease)
   - adds new nucleotides in short fragments called okazaki
   - leaves gaps
   iii. Enzyme DNA ligase fills the gaps
6. Bacteria?

Question 2

Transcription

Answer 2

Transcription: coping of a gene 
- mRNA: new
- rRNA and tRNA are recycled
Copying of a gene: 
- template is only 1 strand
- RNA is only 1 strand 
Site of Transcription
- prokaryote: cytoplasm
- eukaryote: nucleus
- protein synthesis in the ribosome/cytoplasm

Question 3

RNA Splicing

Answer 3

Exon - Intron- Exon - Intron
RNA takes out intron
Exon Exon
Exon - Exon

Question 4

Translation

Answer 4

Creates new protein molecules
Central Dogma: the process of transcription and translation
1. mRNA carries the coes. A copy of 1 strand of DNA in triplets of ACGTU
2. Ribosomes: with rRNA, forms the site for decoding
3. tRNA: brings codes to the factory - delivery code
- ~ 20 amino acids are used universally
- UAC - MET = methionine
4. Genetic code:
- nucleotide sequence, gene in the DNA
- relates to the nucleotide sequence into mRNA
- amino acids in the protein
Combinations
4x4x4=64
- 1 universal start codon
- 3 stop codon
- 20 universal amino acids
= 40 extra codons
Several amino acids have more than 1 codon

Question 5

Codons

Answer 5

Start: AUG
Stop: UAA, UAG, UGA