Exam 2: Microbial genetics Flashcards
Genetics
Study of inheritance and inheritable traits
Genome
The entire genetic complement of an organism
Ploidy
The number of sets of chromosomes in the nucleus of a biological cell
Genotype
Set of genes in an organism
Phenotype
Physical and function traits of an organism
Transcription
The process by which genetic information represented by a sequence of DNA nucleotides is copied into newly synthesized molecules of RNA, with the DNA serving as a template
Translation
The process by which a sequence of nucleotide triplets in an mRNA molecule gives rise to a specific sequence of amino acids during synthesis of a polypeptide or protein
Polycistronic
A description of mRNA that encodes for multiple polypeptides
Compare the structure of prokaryotic and eukaryotic genomes
Prokaryotic
Chromosome are main portion of DNA with associated proteins and RNA
Prokaryotic cells are haploid (single chromosome copy)
Typical chromosome is circular molecule of DNA in nucleoid
Have plasmids…
Small molecules of DNA that replicate independently
Not essential for normal metabolism, growth, or reproduction
Can confer survival advantages
Conjugation
resistance factors
Virulence plasmids
Eukaryotic
Nuclear chromosomes are linear and sequestered within nucleus
Typically have more than one chromosome per cell
Eukaryotic cells are often diploid (two chromosome copies) – up to 1024
Discuss the central dogma of genetics; DNA replication, transcription & translation. Discuss the properties of DNA that facilitate these processes. Describe these processes
Draw out central dogma
DNA replication:
- Starts at the origin of replication (ori)
- DNA polymerase replicates DNA only 5’ to 3’ (nucleotides are added to 3’ end)
- Key to replication is reverse complementary structure of the two strands
- Replication is semiconservative
Transcription:
- Information in DNA is copied as RNA
- Begins at a region of DNA called a promoter (recognized by RNA polymerase) and ends with a sequence called a terminator
- Initiation, Elongation, Termination
Translation:
- RNA → polypeptides, starts AUG
- Ribosomes use genetic information of nucleotide sequences to synthesize polypeptides
- For a new round of peptide chain elongation to occur, the P-site tRNA must move to the E-site and the A-site tRNA must move to the P-site.
Gene
A hereditary unit of information
Discuss the structure of DNA
-Strands are reverse complementary
-Base pairs: A-T, G-C , NO uracil present
-Double helical
-Runs 5’ to 3’
-DNA grows from the 3’ end (adding nucleotides to the 3’ end)
-Held together by hydrogen bonds between base pairs
-Nucleotides are the monomers that make up NA
-Nucleic acid structure:
1. Phosphate
2. Pentose sugar (deoxyribose of ribose)
3. 1 of 5 cyclic nitrogen bases
- Pyrimidines: TCU
- Purines: A&G
Compare and contrast inducible operons and repressible operons
Inducible operons: must be activated by inducers
- Lactose operon
- Catabolic pathways → presence of substrate
Repressible operons: transcribed continuously until deactivated by repressors
- Tryptophan operon
- Anabolic pathway → presence of product
strand codes
5’ CGATGCGTAGC 3’ (coding strand bc it looks like the mrna code)
So opposite strand
3’GCTACGCATCG 5’ ( template strand bc it is the template the rna polymerase uses)
mRNA
5’CGAUGCGUAGC 3’
5’CG AUG (methionine)CGU (arginine) AGC (ser) 3’
Mutations
Mutations
There are multiple types of repair systems but its once the replication process as occurred
-Spontaneous mutations
Slippage leading to an insertion and it shifts
-Induced mutation
Directly damages the DNA
Don’t call base pair substitution a point mutation
-Silent - if it stays a proline it is silent
-Missense - change from pro to cys
-Nonsense = stop codon
