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68 Cards in this Set

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DNA structure
double helix

side =sugar and phospate back bone

inside = hydrogen bonded
nitrogenous bases

width = 2nm

length per circle = 3.4 nm
10 base pair = 10 nucleartide
nucleotide structure

difference btw DNA and RNA
phosphate-sugar-nitrogenous base

in sugar
DNA has H on the c next to the c connected to nitrogen base

RNA has OH

Phospodiester Bond (btw P and S)

Glycosidic Bond (btw S and N)

nucleotide is added to 3 hydroxyl group

Nitrogen base
-pyrimidines
--cytosine, thymine(in DNA), uracil(in RNA)

-purine
--guanine
--adenine
Phospodiester Bond
in nucleotide

btw P and S
glycosidic bond
in nucleotide

BTW S and N
Nitrogen Base
-pyrimidines
--cytosine, thymine

-purine
--guanine
--adenine

cyosine:::guanine (3 H bond)

thymine (uracil in RNA)::adenine
CUT the PYE

Cytosine, Uracile, thymine are pyrimidines
Chargaff Rule
A::T(double bond) and C:::G(triple bond) 양이 같다
DNA replication
semi-conservative (parent strand is paired with daughter strand)

daughter strand is made from 5->3
leading strand 5->3 (thus stick to 3 end of parent strand)

Okazaki fragment stick to 5 end of parent strand but replicates by fragment from 5->3 and Ligase connects the frgments

1. DNA is unwind from the histone protein by DNA helicase

2. Break is created in the DNA by topoisomerase which also stablize broken strand

3. DNA POL comes in and replicate
-origin of replication
--Prok : one
--euk: many
Helicase
double helix 가 풀어진다
SSB protein
helicase 에 의해 풀린 가닥이 결합하는걸 방지
Primase
RNA Primer를 만든다
DNA Pol III
can't start w/o RNA primer
parent strand 3end 애 새로들어오는 nucleotide 를 붙인다
DNA POL I
RNA primer 를 5 말단으로부터 제거 RNA nucleotide 로 대치
Ligase
connect lagging strand (Okazaki fragment)
Gyrase
DNA 엉킴방지
topoisomerase I and II
I : DNA 한가닥 절단
II: DNA 두가닥 절단

also stablize the broken strand
Prokaryote DNA POL
I, II, III

I III proof read
I exonuclease base 제거
eukaryototic DNA POL
alpha, beta, Y, sigma,E

Alphal: lagging strand
sigma: leading strand
types of RNA
mRNA: messenger RNA, made in nucleus

tRNA: transfer RNA, bring amino acid to ribosome, made in cytosol

rRNA: ribosomal RNA;component of ribosome, made in nucleoli
tRNA
each tRNA is specific for one a.a

has two ends, one end has anticodons, other end has a.a that is corresponds to codon
Transcription
3 step
-initiation
-elongation
-termination
process of making mRNA

like replication mRNA is made from 5 -->3

in nucleus

DNA cannot leave the nucleus or it will be cleaved by cytoplasmic DNAse

helicase-Topiosomerase-RNA POL
Transcription

Initiation
start from the Promotor (TATAA BOX)

Function of Promotor
-determine where to begin
-determine which strand is template

RNA POL 은 DNA promotor 를 인식
Function of Promotor
Function of Promotor
-determine where to begin
-determine which strand is template
Transcription

Elongation
RNA POL 은 20 b.p 씩 DNA의 base H-bond 를 끊으면서 unwinds DNA

No proof reading (unlike DNA POL)

5-->3 direction of mRNA
Transcription

Termination
RNA POL 은 Terminator 라는 염기서열에 도착하면 종결
Post transcriptional RNA processing
RNA (intron/exon)splicing

addition of methyguanine Cap
(capping)

Roly A tali

prevents mRNA from degrading as it leaves the nucleus
RNA splicing
DNA 에서 온 intron (noncoding zone) 을 컷, exon (coding zone) 을 이어준다
capping
5 end 에 7 methyguanosine 이 결합

protect mRNA from hydrolytic enzyme

ribome이 mRNA 를 인식하는 신호작용

핵에서 세포질로 mRNA 이동증진
poly A tail
3end 에 AAUAAA sequence

mRNA 분해돼는걸 방지

mRNA 번역 증진
Prokaryote DNA POL
I, II, III

I III proof read
I exonuclease base 제거
eukaryototic DNA POL
alpha, beta, Y, sigma,E

Alphal: lagging strand
sigma: leading strand
translation

initiation
elongation
termination
RNA to Protein

initiation begins when tRNA complex attaches to the mRNA at the start codon (AUG)

Termination take place when the tRNA complex reaches one of the stop codon (UAA, UAG, UGA)
Start codon
AUG
Stop codon
UAA, UAG, UGA
Translation

initiation
small ribosomal subunit가 mRN의 5' 말단에 있는 leader's segment와 결합
(prokaryotes, bacteria: shine delgado sequence 와 결합)

(euk: AUG start codon)

MET (a.a) which codes for AUG 를 운반해온 tRNA 가 mRNA 에 붙는다 (Psite)

그리고 large subunit 가 붙는다.
Translation

elongation
a.a 는 첫번째 a.a 뒤에 하나씩 붙는다.

1st codon recognition (after start codon)

peptide formation
-P 자리에 tRNA에 있는 A.A 가 A 자리 tRNA 와연결

translocation
P to E site
A to P site
Large subunit

binding Site
A: aminocyl tRNA binding site

P: peptidyl tRNA binding site

E: exit site
Translation

Termination
elongation until stop codon reaches at A site

release factor 가 A site stop codon 에 결합
-polypeptide chain 을 tRNA로부터 분해 (hydrolysis)
-large/small subunit 분해
protein synthesis
occurs at ribosome

small subunit attach to mRNA

Large subunit attach to tRNA
tRNA + a.a by??
aminoacyl tRNA synthetase
# of codon greater than a.a ??

degeneracy(redundancy) of the genetic code
yes

thus a.a 는 하나 이상의 코돈에서 암호화된다.

64 codons> 20a.a thus a.a represented by more then one codon
protein funtionality
it must adopt a tertiary structure (which is determined by primary structure)
mutation
2 types
germline mutation
-occur in gamte cell
-pass to offspring

somatic mutation
-lead to tumor

mutations are chages in theDNA caused by mutagens: UV, radiation or chemical)
Point mutation
when single nucleotide base (C,U,T,A,G) is subsituded by another base

lead to DNA damage
types of point mutation
silent
missense
nonsense
Toguchi
Silent mutation
point mutation

a change ina single base of mRNA coding strand does not change the a.a that is produced by the codon
due to degeneracy

codon 의 중복성 thus no a.a change
Missense Mutation
point mutation

mutation in the coding strand cause the production of a different protein then intended
Nonsense mutation
point mutation

changed base in the coding strand is code for stop codon which produce non-functional protein
Toguchi mutation
point mutation

due to the redundency, base change code for the protein that was originally intended
Frameshift Mutation
insertion or deletion of multiple nucleotides which results in a shift in the reading frame
(choromosome mutation)

-delete
-duplicate
-inversion
-translocation
Kinefelter syndrome
XXY
extra X choromosme
Down Syndrome
XXX

trisomy of chromosome 21

nondisjuction
-failure of homologus choromosme to seperate properly during meiosis or failure of sister cromatid to seperate properly during meiosis II

-result zygote have threee copies of that chromosome
Turner syndrome
X

only in felmeld
nondisjunction
failure of homologus chromosome to seperate properly during meiosis or failure of sister cromatid to seperate properly during meiosis II

result zygote have three copies of chromosome
operon
gene reglaton system
consist of operator gene and structural gene
gene expresstion (transcription)
regulator gene
promotor gene
operatior gene
structural gene
regulator gene
produce repressor molecule
promotor gene
binding site for RNA POL

non coding
Operator gene
is the sequence of nontranscribable DNA that is repressor binding site
structural gene
contain sequence of DNA that code for proteins
inducible system
repressor binds to the operator and prevent RNA POL from transcribing the structual gene

inducer binds to the repressor and form inducer repressor complex and can't bind to operator and transcribe occur

inducer is usually substrate an structureal gene code for enzyme
thus when substrate is there then enzyme is produced
repressible system
repressor can't bind to the operator and prevent transcription

need corepressor (end product)

repressor corepressor complex bind to the operator and stop transcription

protein produced by repressible sys.
tRNA molecules have the a.a bound to the () end of the moelecule
3 end
during protein synthesis message is read from?
5 to 3
protein synthesis (translation) require energey

true or false?
true
prokaryotes do not have () organelle
membrane bound thus do not have neucleus
erythrocyte (red blood cell)does not have what??
DNA because they do not have neucleus so that they can store more hemoglobin in humans and other mammals
lac operon
is set of cntrol and structural gene in E Coli. that allow digeston of lactose

follow inducible system

lack of lactose a repressor protein is bound to the operator, preventing RNA POL from binding to the DNA thus preventing translation of the structural gene

when lactoe is present and glucose is absent the repressor is removed and RNA POL can attach to promotor and translation occur
must know the difference when question is asking prabability genetic question
and when question is askng ration of phenotype.
example of prabability genetic question is q7 pcat bio 1