DNA Structure Flashcards
DNA nucleotides
consist of a pentose sugar with 3 important groups attached
1’ = nitrogenous base
3’ = OH group
5’ = 3 phosphate groups
complementary base pairing
hydrogen bonds hold together complementary bases
A/T = 2
G/C = 3 (requires more heat to break)
Chargaff’s Rule
the amount of adenine is equal to the amount of thymine and the amount of guanine is equal to the amount of cytosine
the amount of pyrimidines is equal to the amount of purines
TRUE due to the base pairing properties
Watson and Crick’s Model
sugar backbone on the exterior with nitrogenous base pairing occurring in the interior
two DNA strands are antiparallel
sugar backbone
created via phosphodiester bonds between the 3’ hydroxyl and the 5’ phosphate of two bases
pyrophosphate cleaved provides energy for bond to form
can only add bases to the 3’ end because there is a free hydroxyl and the 5’ carbon only has one phosphate so no energy can be released
conservative
parental DNA is denatured, each strand is used as a template to create a new strand, parental DNA binds back together and new strands bind
parental DNA maintains its identity
semi-conservative
parental DNA denatures, each strand is used as a template to create a new complementary strand, maintains this association
each parental strand maintains its identity
dispersive
parental strands denature and fragment into pieces that act as templates for small portions of new DNA
parental strands DO NOT maintain identity
Meselson and Stahl control experiment
e. coli was grown one heavy nitrogen and light nitrogen then centrifuged
light nitrogen was higher in the tube than heavy nitrogen
established that it is possible to separate DNA out by the average weight of the nitrogen it contains
Meselson and Stahl experiment
DNA was replicated on heavy nitrogen for a few rounds then moved to a light medium
first generation: intermediate band (semi-conservative or dispersive)
second generation: intermediate and light band (semi-conservative)
PCR
method of studying DNA and performing genetic manipulations
reaction is repeatedly heated and cooled
PCR requirements
DNA sample/target sequence
DNA nucleotides
synthetic DNA primers
DNA polymerase that can withstand high temperatures during the denaturing process
target sequence
piece of DNA that comes from a sample that is copied to study further
primers
short sequences of DNA that are synthesized in the lab
complementary to the 3’ strand of each parental strand
needs to be upstream of the target sequence = makes up the 5’ strand of the new DNA
hot and cool stages of PCR
hot stage: DNA denatures, producing single-stranded DNA
cold stage: primers anneal to the single-stranded DNA at the sites where they match, DNA polymerase builds a new strand starting at each primer, last long enough so that DNA polymerase can build complementary DNA
PCR equation
N = No x 2^x
