Bio Subject Test 1 And 2 "molecular Genetic"

Front 1

DNA structure

Back 1

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

Front 2

nucleotide structure

difference btw DNA and RNA

Back 2

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

Front 3

Phospodiester Bond

Back 3

in nucleotide

btw P and S

Front 4

glycosidic bond

Back 4

in nucleotide

BTW S and N

Front 5

Nitrogen Base

Back 5

-pyrimidines
--cytosine, thymine

-purine
--guanine
--adenine

cyosine:::guanine (3 H bond)

thymine (uracil in RNA)::adenine

Hint 5

CUT the PYE

Cytosine, Uracile, thymine are pyrimidines

Front 6

Chargaff Rule

Back 6

A::T(double bond) and C:::G(triple bond) 양이 같다

Front 7

DNA replication

Back 7

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

Front 8

Helicase

Back 8

double helix 가 풀어진다

Front 9

SSB protein

Back 9

helicase 에 의해 풀린 가닥이 결합하는걸 방지

Front 10

Primase

Back 10

RNA Primer를 만든다

Front 11

DNA Pol III

Back 11

can't start w/o RNA primer
parent strand 3end 애 새로들어오는 nucleotide 를 붙인다

Front 12

DNA POL I

Back 12

RNA primer 를 5 말단으로부터 제거 RNA nucleotide 로 대치

Front 13

Ligase

Back 13

connect lagging strand (Okazaki fragment)

Front 14

Gyrase

Back 14

DNA 엉킴방지

Front 15

topoisomerase I and II

Back 15

I : DNA 한가닥 절단
II: DNA 두가닥 절단

also stablize the broken strand

Front 16

Prokaryote DNA POL

Back 16

I, II, III

I III proof read
I exonuclease base 제거

Front 17

eukaryototic DNA POL

Back 17

alpha, beta, Y, sigma,E

Alphal: lagging strand
sigma: leading strand

Front 18

types of RNA

Back 18

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

Front 19

tRNA

Back 19

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

Front 20

Transcription
3 step
-initiation
-elongation
-termination

Back 20

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

Front 21

Transcription

Initiation

Back 21

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

Front 22

Function of Promotor

Back 22

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

Front 23

Transcription

Elongation

Back 23

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

Front 24

Transcription

Termination

Back 24

RNA POL 은 Terminator 라는 염기서열에 도착하면 종결

Front 25

Post transcriptional RNA processing

Back 25

RNA (intron/exon)splicing

addition of methyguanine Cap
(capping)

Roly A tali

prevents mRNA from degrading as it leaves the nucleus

Front 26

RNA splicing

Back 26

DNA 에서 온 intron (noncoding zone) 을 컷, exon (coding zone) 을 이어준다

Front 27

capping

Back 27

5 end 에 7 methyguanosine 이 결합

protect mRNA from hydrolytic enzyme

ribome이 mRNA 를 인식하는 신호작용

핵에서 세포질로 mRNA 이동증진

Front 28

poly A tail

Back 28

3end 에 AAUAAA sequence

mRNA 분해돼는걸 방지

mRNA 번역 증진

Front 29

Prokaryote DNA POL

Back 29

I, II, III

I III proof read
I exonuclease base 제거

Front 30

eukaryototic DNA POL

Back 30

alpha, beta, Y, sigma,E

Alphal: lagging strand
sigma: leading strand

Front 31

translation

initiation
elongation
termination

Back 31

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)

Front 32

Start codon

Back 32

AUG

Front 33

Stop codon

Back 33

UAA, UAG, UGA

Front 34

Translation

initiation

Back 34

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.

Front 35

Translation

elongation

Back 35

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

Front 36

Large subunit

binding Site

Back 36

A: aminocyl tRNA binding site

P: peptidyl tRNA binding site

E: exit site

Front 37

Translation

Termination

Back 37

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

Front 38

protein synthesis

Back 38

occurs at ribosome

small subunit attach to mRNA

Large subunit attach to tRNA

Front 39

tRNA + a.a by??

Back 39

aminoacyl tRNA synthetase

Front 40

# of codon greater than a.a ??

degeneracy(redundancy) of the genetic code

Back 40

yes

thus a.a 는 하나 이상의 코돈에서 암호화된다.

64 codons> 20a.a thus a.a represented by more then one codon

Front 41

protein funtionality

Back 41

it must adopt a tertiary structure (which is determined by primary structure)

Front 42

mutation
2 types

Back 42

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)

Front 43

Point mutation

Back 43

when single nucleotide base (C,U,T,A,G) is subsituded by another base

lead to DNA damage

Front 44

types of point mutation

Back 44

silent
missense
nonsense
Toguchi

Front 45

Silent mutation

Back 45

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

Front 46

Missense Mutation

Back 46

point mutation

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

Front 47

Nonsense mutation

Back 47

point mutation

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

Front 48

Toguchi mutation

Back 48

point mutation

due to the redundency, base change code for the protein that was originally intended

Front 49

Frameshift Mutation

Back 49

insertion or deletion of multiple nucleotides which results in a shift in the reading frame
(choromosome mutation)

-delete
-duplicate
-inversion
-translocation

Front 50

Kinefelter syndrome

Back 50

XXY
extra X choromosme

Front 51

Down Syndrome

Back 51

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

Front 52

Turner syndrome

Back 52

X

only in felmeld

Front 53

nondisjunction

Back 53

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

Front 54

operon

Back 54

gene reglaton system
consist of operator gene and structural gene

Front 55

gene expresstion (transcription)

Back 55

regulator gene
promotor gene
operatior gene
structural gene

Front 56

regulator gene

Back 56

produce repressor molecule

Front 57

promotor gene

Back 57

binding site for RNA POL

non coding

Front 58

Operator gene

Back 58

is the sequence of nontranscribable DNA that is repressor binding site

Front 59

structural gene

Back 59

contain sequence of DNA that code for proteins

Front 60

inducible system

Back 60

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

Front 61

repressible system

Back 61

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.

Front 62

tRNA molecules have the a.a bound to the () end of the moelecule

Back 62

3 end

Front 63

during protein synthesis message is read from?

Back 63

5 to 3

Front 64

protein synthesis (translation) require energey

true or false?

Back 64

true

Front 65

prokaryotes do not have () organelle

Back 65

membrane bound thus do not have neucleus

Front 66

erythrocyte (red blood cell)does not have what??

Back 66

DNA because they do not have neucleus so that they can store more hemoglobin in humans and other mammals

Front 67

lac operon

Back 67

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

Front 68

must know the difference when question is asking prabability genetic question
and when question is askng ration of phenotype.

Back 68

example of prabability genetic question is q7 pcat bio 1