Dna Structure, Replication, Translation, Transcription, Gene Expression (Biochem, Cb) Review

Front 1

Enzymes for eukaryotic replication

Back 1

> DNA polymerase delta

=5'>3'

=3'>5' exonuclease

> DNase I

= 3'>5' exonuclease

> RNA primase

> DNA ligase (fill in 'nicks')

> HELICase

Front 2

Enzymes for prokaryotic replication

Back 2

RMT

> DNA polymerase 1 (rRNA)

> DNA polymerase 2 (mRNA)

> DNA polymerase 3 (tRNA)

= 5'>3'

=3'>5' exonuclease

---

> helicase

> DNase I

Relieving supercoiling, stress

> Topoisomerase IV (gram +) (- quinolone)

> DNA gyrase (gram -) (quinolone, i.e: ciprofloxacin)

Front 3

Types DNA repair mechanisms

Back 3

Nucleotide excision repair

- Disorder causes xeroderma pigmentosum

steps

> DNA endonuclease

> DNA polymerase, ligase

Base excision repair

> DNA glycosylase

> DNA endonuclease

> DNA polymerase, ligase

DNA mistmatch repair

- Disorder causes Hereditary nonpolyposis colorectal cancer (aka, Lynch syndrome)

Mechanism:

> Nonmethylated DNA strand (new strand) is worked on

Front 4

DNA disorders

Back 4

Thymine dimer

> cause: UV light

Guanine adduct

> cause: oxidation of Cigarette smokg

Xeroderma pigmentosum

> cause: faulty nucleotide excision repair

Tx: dimericine

Hereditary nonpolyposis colorectal cancer

> cause: faulty DNA mismatch repair

Front 5

Dimericine is used to treat what?

a. thymine dimer

b. xeroderm pigmentosum

c. heredity nonpolyposis colorectal cancer

d. guanine adduct

e. ATM mutation

f. ATR mutation

Back 5

b. xeroderma pigmentosum

Expl: restores the nucleotide excision repair mechanism that is faulty. in xeroderm pigmentosum

Front 6

A defect in nucleotide excision repair causes what?

a. thymine dimer

b. xeroderm pigmentosum

c. heredity nonpolyposis colorectal cancer

d. guanine adduct

e. ATM mutation

f. ATR mutation

Back 6

a. thymine dimer

b xeroderm pigmentosum

d. guanine adduct

Front 7

A defect in mismatch repair causes which of the following?

a. thymine dimer

b. xeroderm pigmentosum

c. heredity nonpolyposis colorectal cancer

d. guanine adduct

e. ATM mutation

f. ATR mutation

Back 7

c. heredity nonpolyposis colorectal cancer

Expl: A defect in mismatch repair causes hereditary nonpolyposis colorectal cancer.

Front 8

Structure nucleic acid

Back 8

Nucleotide polymers

Nucleoside = N2 base + sugar

Nucletide = Nucleoside + PO43- = N2 base + sugar + PO43-

1' attached to base VIA N-glycosidic bond

2': -H in DNA, -OH in RNA (Makes it less stable)

3': attached to PO43-

5': attached to PO43-

Phosphodieste rbackbone

Front 9

Bases connect to sugar via what?

Back 9

N-glycosidic bond

Front 10

Charge of DNA?

Back 10

- due to negative charges of pO43- in the sugar phosphate backbone

Front 11

In DNA, th bond between deoxyribose sugar + phosphate best described by what?

a. polar bond

b. ionic bond

c. hydrogen bond

d. covalent bond

e. van der waals bond

Back 11

d. covalent bond

expl: phopshodiester bond is a type of covalent bond

Front 12

Explain base pairing

Back 12

A (2 H2 bonds) T

G (3 H2 bonds) C

Charfaff's rules: [A] = [T], [G] = C] or purines = pyrimidines

Front 13

Explain Tm

Back 13

(denaturation temp or temp at which 50% strands denatured) = 69.3 + 0.41 (%G-C)

the %G-C is a WHOLE number, not a decimal

Front 14

Impact of adding basic solution to nucleic acid.

Back 14

Only denature DNA

Breaks apart RNA, b/c ribose has a 2-OH that deprotonates in basic solution and then breaks apart phopshodiester bond.

Front 15

Explain nucleic acid hybridization

Back 15

Single strands with complementary sequences can hybridize, which can be useful in molecular biology research.

Front 16

Explain stability between bases, why are bases stable in DNA?

Back 16

Pi stacking interactions between bases to increase stability

Front 17

Match.

1. Dominates in living systems and is right handed

2. Dominates in DNA-RNA hybrids, "tightly wrapped."

3. Dehydration form of B-DNA

4. Line connecting phosphate zags

a. A DNA

b. B DNA

c. Z DNA

Back 17

1. B-DNA

2. A-DNA

3. A-DNA

4. Z-DNA

Front 18

Describe chromatin and its structure

Back 18

Chromatin = DNA + various proteins

Beads on a string essentially: complexes of histone proteins + DNA = centrosomes

Front 19

What are centrosomes?

Back 19

Complexes of histone proteins (2 H2A, 2 H2B, 2 H3, 2 H4) octamer w/ DNA

Front 20

Describe histones

-structure

-types

Back 20

Structure: High % of lysine and arginine that are able to interact w/ - charged sugar phosphate DNA backbone

~ 140 base pairs of DNA wrap around each nucleosome core

Types histones:

8 on nucleosome

> 2 H2A, 2 H2B, 2 H3, 2 H4

H1 outside between the nucleosomes 'beads'

Front 21

How many base pairs of DNA wrap around nucleosome core?

Back 21

About 140 base pairs.

Front 22

What is a gene?

Back 22

Specific point on chromosome that has a functional role in producing a protein

Front 23

Explain 5-Fluouracil (5-FU)

- Use

- Mechanism

Back 23

Normally uracil - methylation at C-5 --> 5-methyl uracil

But the 5-FU inhibits thimidylate synthase, inhibiting production of deoxythymidine.

Use: slowing progression of tumor cell

Front 24

Azithromycin

Ciprofloxazin

Back 24

Azithro

> binds 50S ribosome subunit, preventing protien synthesis

Ciprofloxacin

> targets bacterial DNA gyrase

Front 25

Explain retroviruses

Back 25

Have RNA genomes

Use enzyme, reverse transcriptase (RT) tomake DNA copy to insert into nucleus of host cell

Front 26

Compare ribosomes + rRNA vs pro + euks

Back 26

Pros/euks:

ribosome = 30s, 50s, 70s / 40S, 60S, 80S

rRNA = 3 types rRNA (23, 16, and 5S) / 4 types rRNA (28, 18, 5.8, 5S)

Front 27

Explain proteins in prokaryotic replication

Back 27

Helicase

> unwind helix to separate pairs

Topoisomerase, gyrase

> rleive supercoiling

Single stranded binding protien

> maintains single strands

Front 28

Match.

1. What is involved in relieving supercoiling?

2. What unwinds helix to separate pairs before DNA replication?

3. What maintains single strands?

4. Ciprofloxacin inhibits what?

a. DNA gyrase

b. helicase

c. single stranded binding protein (SSB)

d. topoisomerase

Back 28

1. DNA gyrase, topoisomerase

2. DNA helicase

3. single stranded binding DNA

4. ciprofloxacin inhibits DNA gyrase

Front 29

Describe DNA polymerase activity

Back 29

> Primes adds RNA primer

= Reason: DNA polymerase needs 3' OH to function to perform nucleophilic attack

> Works 5'>3' directions

> Adds new deoxynucleotide triphosphate to 3' OH group of chain while releasing PPi, (very exergonic)

> Primers then removed by RNase H and DNA polymerase I (via DNA polymerase 1's 5'>3' exonuclease activity)

> Okazaki fragments ( on lagging strand) , joined by DNA ligase

Front 30

1. How are primers removing during DNA synthesis?

2. Okazaki fragments are joined by what?

a. DNA polymerase I

b. DNA P 2

c. DNA P 3

d. DNA gyrase

e. RNase H

f. DNA ligase

g. DNA polmyerase delta

Back 30

1.

a. DNA polymerase 1

> via its 5'>3' exonuclease activity

e. RNase H

2. f. DNA ligase

Front 31

Proofreading activities in DNA replication

Back 31

DNA p 1

> 5'>3' exonuclease (removes RNA priemrs)

DNA p 3

> 3'>5' exonuclease activity

Front 32

What stage of mitosis does synthesis occur?

Back 32

S phase

Front 33

Difference between eukaryotic and prokaryotic DNA replication?

Back 33

Prokaryotic:

> DNA Pol I, Replication, 5'>3' exonuclease activity

> Polmyerase III, Major replicative polymerase, 3'>5' exonuclease for proofreading

Eukaryotes

> DNA polymerase delta: replication, 3'>5' exonuclease for proofreading

Diferences: (pro/euk)

> single origin of replication/multiple origins of replication

> similar enzymes (topoisomerase, primes, polymerase, leading lagging strands)

> circular/linear

Front 34

1. What is the major replicative enzyme in prokaryotes?

2. What is the replicative enzyme in prokaryotes with 5'>3' exonuclease activity?

3. What is the replicative enzyme in eukaryotes with 5'>3' exonuclease activity?

4. What is the replicative enzyme in prokaryotes with 3'>5' exonuclease activity?

a. DNA polymerase I

b. DNA P 2

c. DNA P 3

d. DNA gyrase

e. RNase H

f. DNA ligase

g. DNA polmyerase delta

Back 34

1. DNA P 3

> Also has 3'>5' activity

2. DNA P 1

3. None (not listed on slide)

4. DNA P delta

Front 35

What are the ends of chromosomes?

Back 35

Teloemres

Front 36

What enzyme maintains telomere length during replication?

Back 36

Telomerase

Front 37

Teloemrase

> function

> mechanism

> properties

Back 37

Function:

Enzyme that maintains telomere length during replication

Mechanism:

> RNA reverse transcriptase mechanism

Properties:

> Importance in senesence, aging, cancer

> Telomerase may increase in some cancer cells

Front 38

DNA methylation

-result

-mechanism

-importance

Back 38

Result:

> Inhibits gene expression

Mechanism

> Adenine, cytosine methylated

Importance

> gene regulation

> epigenetics

Front 39

1. DNA mutations due to cigarette smoking that are uncorrected is a result in a defect of which of the following?

a. nucleotide excision repair

b. DNA mismatch repair

c. DNA base excisino repair

d. ligation

2. The product resulting from the effects of cigarette smoking on DNA reacts with what to distort DNA helix?

a. adenine

b. guanine

c. thymine

d. cytosine

e. uracil

3. What is a possible outcome of this?

a. malignant melanoma

b. brain cancer

c. adenocarcinoma

d. pyrmidine dimer

e. hereditary nonpolyposis colorectal cancer

Back 39

1. a. nucleotide excision repair

2. b. guanine

> guanine adduct is formed

3. c. adenocarcinoma

Front 40

DNA mutations due to UV light exposure that are uncorrected can result in what?

a. malignant melanoma

b. brain cancer

c. adenocarcinoma

d. pyrmidine dimer

e. hereditary nonpolyposis colorectal cancer

Back 40

a. malignant melanoma

Front 41

DNA mutations due to UV light exposure result in what usually?

a. malignant melanoma

b. brain cancer

c. adenocarcinoma

d. pyrmidine dimer

e. hereditary nonpolyposis colorectal cancer

Back 41

d. poyrimidine dimer

> later can become malignant melanoma if not treated

Front 42

A person has xeroderma pigmentosum. This is due to a defect in which of the following?

a. nucleotide excision repair

b. DNA mismatch repair

c. DNA base excisino repair

d. ligation

Back 42

a. nucleotide excisionr epair

Front 43

A person has a defect in DNA mismatch repair mechanisms. This leads to what?

a. malignant melanoma

b. brain cancer

c. adenocarcinoma

d. pyrmidine dimer

e. hereditary nonpolyposis colorectal cancer

Back 43

e. hereditary nonpolyposis colorectal cancer

Front 44

Matching

1. Which occurs during meiosis?

2. What are 'jumping genes' or mobile genetic elements?

3. Which results in Holliday structure formation?

4. What is caused by breaks in 2 non homologous chromosomes?

5. What uses reverse transcription to make a DNA copy?

a. homologous recombination

b. retroviral insertion

c. translocation

d. transposons

Back 44

1. a. homologous recombination

2. d. transposons

3. a. homologous recombination

4. translocation

5. retroviral insertion

Front 45

Deaminations and significance

Back 45

Deamination of cytosine to uracil can be repaired

Deamination of 5-methylcytosine cannot be repaired.

Front 46

1. Defect in one of several mismatch repair genes result in what?

2. Which of the following may progress to malignant melanomas?

3. Which of the following can cause an adenocarcinoma?

4. Defect in nucleotide excision repair results in what?

5. Dimericine is a treatment for what?

a. xeroderma pigmentosum

b. hereditary nonpolyposis colorectal cancer

c. thymine dimer

d. guanine adduct

Back 46

1. hereditary nonpolyposis colorectal cancer

2. thymine dimers

3. guanine adduct

4. xeroderma pigmentosum, thymine dimer, guanine adduct

5. Xeroderma pigmentosum

Front 47

How is RNA polymerase difference from DNA polymerase?

Back 47

No 3'>5' exonuclease (as in DNA polymerase 1), but both 5'>3' ok

Front 48

1. What produces tRNA?

2. What produces mRNA?

3. What produces rRNA?

4. What produces 5S rRNA?

5. Alpha amantin inhibits what?

a. RNA polymerase I

b. RNA polymerase II

c. RNA polymerase III

Back 48

rmt/5S rRNA

1. RNA P 3

2. RNA P 2

3. RNA P 1

4. RNA P 3

5. RNA P 3

Front 49

Pro vs. euk transcription

Back 49

Pro/euk:

type mRNA:

polycystronic mRNA (1 mRNA many proteins)/ monocystronic mRNA (1 mRNA, 1 protein)

starting factors:

sigma factor/TFIID

connections:

transcription, translation together/translation, transcription separated

location transcription:

cytosol/nucleus

Front 50

1. Redunancy and degeneracy occurs during what?

a. DNA replication

b. transcription

c. translation

Back 50

1. translation

Front 51

1. Putting in the wrong AA is an example of what type of mutation?

2. Prematurely ending translation is due to what type of mutation?

3. Cystic fibrosis is an example of what type of mutation?

4. Progeria is due to what type of mutation?

5. Tay sachs disease is due to what type of mutation?

a. silent

b. missence

c. nonsense

d. insertion

e. deletion

Back 51

1. missence

2. nonsense

3. missence mutation

4. Silent mutation

5. Insertion (Biochem dan , slide 171)

Front 52

What is the first AA in translation?

Back 52

Euks: methionine

proks: formyl-methionine

Front 53

1. Transcription of what gene results in the breakdown of lactose into glucose and galactose

2. Transcription of what gene results in transport of of lactose into cells?

3. Transcription of what gene result in acetylation of B-galactosides?

a. lac a gene

b. lac x gene

c. lac y gene

d. lac z gene

Back 53

1. lac z gene (B-galactosidase)

2. lac y gene (lac permease)

2. lac a gene (transacetylase)

Front 54

Explain positive and negative control of lac operon?

Back 54

Negative: lactose present, lactose binds to receptor and allows RNA polymerase to bind

Positive: no glucose, CAP (bound activator) protein binds to activator site for binding protein

Net result: transcription of lac operon with no glucose and lactose present

Front 55

Explain tryptophan and attenuation

Back 55

Without tryptophan

Result: Transcription occurs of tryptophan receptor

Mechanism: segments 2, 3 bind allowing slower transcription

With tryptophan (attenuation results)

Result: Transcription of tryptophan receptor doe snot occur

Mechanism: segments 3, 4 bind allowing faster transcription and thus limited production tryptophan receptors

Front 56

Heterochromatin to eucharomatin

Back 56

Acetylation allows activation of DNA (heterochromatin to euchromatin)

Front 57

DNA methyulation experiments on RAT

Back 57

DNA methyulation: gene deactivation > high rat anxiety > no care of young

No DNA methyulation: gene activation remains > low rat anxiety > grooming + care of young