[BIOL 1510] - Midterm Exam Guide - Comprehensive Notes for the exam (29 pages long!)

80 views29 pages
7 Feb 2017
Department
Course
Professor

For unlimited access to Study Guides, a Grade+ subscription is required.

Georgia Tech
BIOL 1510
MIDTERM EXAM
STUDY GUIDE
Unlock document

This preview shows pages 1-3 of the document.
Unlock all 29 pages and 3 million more documents.

Already have an account? Log in
Unlock document

This preview shows pages 1-3 of the document.
Unlock all 29 pages and 3 million more documents.

Already have an account? Log in
Cell Division Cycle:
Cells reproduce genetically identical copies of themselves by cycles ofcell growth and
division
This happens in 4 stages:
G1: period after cell division and before the start of replication
Cells grow and monitor their environment to tell if they should start
another round of replication
S: period of DNA synthesis, where cells replicate their chromosomes
G2: period between the end of DNA replication and the start of cell division
Cells make sure replication is complete & makes necessary repairs
M: period of cell division
P,M,A,T,CK
LO1:
Interphase: (G1+S+G2) chromosomes condense in the form of chromatin and become
visible (by microscopy) as eukaryotic cells enter mitosis or meiosis
Prophase: chromosomes = sister chromatids
Mitosis:
Produces 2 daughter cells, genetically identical to each other and the parent
Cells have 2N chromosomes, but 4X DNA content b/c it has been replicated
IN THE END: Cells have 2N chromosomes & 2X DNA content
All Eukaryotic cells except for sex cells
Stages
Prophase: chromosomes condense, chromatids joined @ centromere
Metaphase: chromosomes line up @ plate, pushed and pulled by microtubules
Anaphase: sister chromatids separate & go toward opposite ends of cell
Telophase: chromosomes cluster at ends of the cell and condense
Cytokinesis: membrane pinches, 2 daughter cells
Meiosis:
Produces haploid gametes from diploid cells
Cells have 2N chromosomes & 4X DNA content
IN THE END: gametes have 1N chromosomes & 1X DNA content
Most genetic action occurs in the 1st division
Stages:
Prophase 1: Homologous chromosomes (tetrads) pair up and align end-to-end,
crossing over occurs
Metaphase 1: Tetrads line up at the metaphase plate
Anaphase 1: Homologous chromosomes seperate but sister chromatids remain
together, they move away from the metaphase plate
Telophase 1: Each cell has one of the replicated chromosomes from each
homologous pair
MEIOSIS 2: identical to mitosis in a haploid cell
find more resources at oneclass.com
find more resources at oneclass.com
Unlock document

This preview shows pages 1-3 of the document.
Unlock all 29 pages and 3 million more documents.

Already have an account? Log in
LO1/2: Describe the Central Dogma of Molecular Biology. Describe the DNA sequence motifs
and proteins required to initiate transcription.
Francis Crick : “Central Dogma” = flow of info from nucleic acid to protein
DNA > RNA > Amino Acids/Proteins
Transcription: copying of genes from DNA in order to make mRNA
DNA does not leave the nucleus
RNA polymerase depends on DNA-binding proteins (“transcription factors”) to
bind to special sequence motifs (“promoters”) to recognize where genes start
Transcription factors recruit RNA polymerase to bind to the promoter & begin
transcription nearby
RNA polymerase reads the template strand at the start of the gene, moving along
the DNA making the mRNA from free bases in the nucleus.
mRNA is then processed (edited) and moves into the cytoplasm
LO3: Describe the process of and key components required for translation.
Translation: process of using mRNA molecule as a template to make a protein
Requires 3 components (charged w/ amino acids):
mRNA - transcribed from protein-coding genes
Ribosomes - large assemblies of ribosomal RNAs & proteins (three sites)
Exit site, peptidal, A- site
tRNA - match amino acid to codon in mRNA
Amino-acyl tRNA synthetases recognize various tRNAs and charge them by
attaching the appropriate amino acid.
2 codons fit inside of ribosome
Initiation:
Begins near the 5’ end of mRNA @ AUG triplet
LSU + SSU + initiator tRNA (methionine) - inside ribosome
Ribosome moves along mRNA 3 bases @ a time from 5’ to 3’ end
Codon Recognition:
1st step of Elongation
tRNAs with complementary anti-codons to mRNA codons arrive w/
amino acids
A peptide bond forms to join the amino acid to the carboxyl end of the
polypeptide chain, tRNA is ejected to make room for a new amino-acyl
tRNA
Polypeptide chains have a free amino group @ N-terminus & a carboxyl group at
C-terminus
New amino acids are added to the C-terminus
Translocation: movement down the chain by one codon
LO4: Predict the likely effects of mutations in DNA on protein amino acid sequence,
structure and function.
find more resources at oneclass.com
find more resources at oneclass.com
Unlock document

This preview shows pages 1-3 of the document.
Unlock all 29 pages and 3 million more documents.

Already have an account? Log in

Get access

Grade+
$10 USD/m
Billed $120 USD annually
Homework Help
Class Notes
Textbook Notes
40 Verified Answers
Study Guides
1 Booster Class