BIOCH Y1 S1: DNA Flashcards
1
Q
4 structural characteristics of DNA
A
- double-stranded helix
- uniform diameter
- right-handed twist
- strands are antiparallel (run in opposite directions)
2
Q
describe the structure of DNA
A
- sugar-phosphate backbones coil around the outside of the helix, bound w/ strong phosphodiester bonds
- nitrogenous bases point towards the centre
- weaker H bonds b/n COMPLEMENTARY nitrogenous bases hold the two strands together (a-t > 2 H bonds and c-g > 2 H bonds)
- phosphate groups link 3’ end of one deoxyribose to the 5’ end of the next
- single strand of DNA has a 5’ phosphate group and a free 3’ -OH group
- coding sequence read 5’-3’
3
Q
why do we have chromosomes
A
- length of human DNA is very big but nucleus is small
- so genome is segmented into chromosomes
4
Q
apart from DNA, what else do chromosomes contain and what is their function?
A
- histone proteins
- physically organise DNA and regulate its activities
5
Q
structure of a chromosome
A
- DNA wraps around 8 histone proteins to form a nucleosome
- nucleosomes fold to form chromatin (50/50 DNA:proteins)
- chromatin condenses to form a chromosome
6
Q
2 types of chromatin
A
- heterochromatin: very tightly wound DNA > harder to access for transcription
- euchromatin: looser DNA > easier to access for transcription
7
Q
how does RNA differ from DNA
A
- ribose instead of deoxyribose (extra 2’ -OH group in ribose)
- single stranded (extra -OH group means it’s more stable as a single strand)
- uracil instead of thymine
8
Q
3 models of DNA replication
A
- conservative: one DNA molecule will be completely new and the parental one will remain
- semi-conservative: each molecule of DNA formed will have one new strand and one parental strand (actual model)
- dispersive: both DNA molecules will be a mix of both a parent strand and a new strand
9
Q
DNA replication process
A
- topoisomerase untwists the double helix ahead of the replication fork
- single-stranded binding (SSB) proteins prevent 2 DNA strands from coming back together
- helicase separates the 2 strands by breaking H bonds b/n complementary bases > each strand acts as a template for a new, complementary strand
- primase adds an RNA primer to the 3’ end of the leading and lagging strand TEMPLATES (3’-5’) which now becomes the 5’ end of the NEW leading and lagging strands
- DNA polymerase III adds free nucleotides to synthesise the NEW leading strand from 5’-3’
- NEW lagging strand is synthesised in Okazaki fragments b/c the 5’ end of the lagging strand is towards the replication fork so it can’t get to the very end > has to keep going back
- RNA primers eventually degraded by exonuclease and replaced by DNA > DNA ligase joins Okazaki fragments together
10
Q
3 common types of DNA damage
A
- deaminated cytosine: -NH2 group removed from C to form U > complementary code will be copied wrong during replication
- bulky lesions (pyridimine): T-T, T-C, C-C (not bonded to A or G)
- double-strand break: sugar-phosphate backbone breakage
11
Q
why do we have telomeres and how are they formed?
A
- @ end of chromosome, there’s no 3’ -OH group to prime DNA synthesis > can’t replicate the end of the chromosome so 3’ overhang
- telomerase contains the complementary RNA code for the telomere which allows it to elongate the 3’ end of the original strand
- rest of lagging strand is synthesised
- telomeres (sequence of DNA repeats) act as a cap on the end to prevent the DNA from getting shorter each time it replicates and essentially prevent premature ageing
12
Q
3 mechanisms to increase DNA replication fidelity (accuracy)
A
- preferential recognition: correct pairing of complementary nucleotides by DNA polymerase III
- exonucleolytic proofreading: corrects errors during replication by backtracking, removing the wrong nucleotide and adding the correct one
- strand-directed mismatch repair: mismatches in double helix structure are detected and excised, gap is filled by DNA polymerase and ligase
13
Q
4 mechanisms of DNA repair
A
- base excision repair: repairs one incorrect nucleotide i.e. because of deamination
- nucleotide excision repair: repairs a few faulty nucleotides i.e. because of the bulky damage
- non-homologous end joining: 2 strands trimmed and then joined i.e. because of the double-strand break (however deletion can have consequences)
- homologous recombination: sister chromatid can be used as a template to repair the damaged chromatid i.e. because of the double-strand break
14
Q
what is an origin site and how is it different in eukaryotes and prokaryotes
A
- where DNA replication begins
- prokaryotes: one ori site on circular plasmid
- eukaryotes: several ori sites on linear DNA
