Exam 3

Question 1

Initiator proteins

Answer 1

breaks H bonds to initiate replication

Question 2

Primer

Answer 2

short nucleotide sequence that initiates DNA polymerase binding

Question 3

Primase

Answer 3

RNA polymerase that creates primers

Question 4

For proofreading to take place…

Answer 4

DNA must be copied in 5 - 3’ direction

Question 5

why is it 5’ to 3’

Answer 5

energy from triphosphate on oncoming 5’ end gives energy for reaction

Question 6

Telomerase

Answer 6

Adds bases 5’ to 3’ to the lagging template strand

*Extends the template strand so DNA replication proteins can replicate
lagging strand and it does not shrink with each cell division.

Question 7

Depurination

Answer 7

G lost

Question 8

Deamination

Answer 8

C turns to U

Question 9

Mismatch repair system

Answer 9

Excision - damage is
recognized and cut out by one of a
series of nucleases

resynthesis - original
DNA sequence is restored by a
repair DNA polymerase

Ligation - DNA ligase seals
the break left in the sugar–
phosphate backbone of the
repaired strand

Question 10

Exonuclease

Answer 10

– remove bases from the end of a strand

Question 11

Endonuclease

Answer 11

– remove bases from the middle/cut DNA or RNA

Question 12

5’-3’ exonuclease activity

Answer 12

• Removing RNA primers made by primase

Question 13

3’-5’ exonuclease activity

Answer 13

• Proofreading

Question 14

Nucleases

Answer 14

– enzymes that remove nucleotides from DNA

Question 15

Promoter v Terminator

Answer 15

Terminator transcribed, Promoted not

Question 16

Promoter sequences at

Answer 16

-10, and -35 nucleotides down from start

Question 17

Eukaryotic vs Prokaryotic regulatory proteins

Answer 17

Eukaryotic - multiple accessory proteins
- 3 RNA poly

Prokaryotic - one sigma factor
- 1 RNA poly

Question 18

RNA poly 2

Answer 18

transcribes all proton coding genes

Question 19

TFIID, TFIIH, and RNA poly II

Answer 19

TFIID - distorts DNA

TFIIH - opens up double helix

RNA Poly II - disassociates from other factors and starts transcription

Question 20

Elongation Factors

Answer 20

Form a wedge on nucleosomes that allow RNA poly II to transcribe through proteins

Question 21

cap and poly A tail

Answer 21

5’ cap Phosphate w/ modified guanine

Poly A tail

Question 22

Effects from capping and polyadenylation

Answer 22

 1) increase the stability of eukaryotic mRNA.
 2) facilitate export to the cytosol
 3) mark the molecule as an mRNA molecule

Question 23

snRNP

Answer 23

recognizes splice sequences

Question 24

U1, U2, U6

Answer 24

U1 recognizes the 5′ splice site

U2 recognizes the lariat branch-point site through
complementary base-pairing.

U6 then “double-checks” the 5′ splice site by displacing
U1 and base-pairing with the same intron sequence itself.

Question 25

RNA synthesis location

Answer 25

membrane-less components in nucleus

Question 26

snRNA and snRNP

Answer 26

small nuclear ribonucleoprotein used as functional unit of spliceosome small nuclear RNA recognizes splice site

Question 27

UUU codes for

Answer 27

phenylalanine

Question 28

tRNAS match ____ to codons

Answer 28

amino acids

Question 29

Aminoacyl-tRNA synthetase charges a tRNA with the correct ______ ____, requires ATP

Answer 29

Amino Acid

Question 30

The genetic code is translated by...

Answer 30

aminoacyl-tRNA synthetases and tRNAs.

Question 31

charging

Answer 31

Each synthetase couples a particular amino acid to the proper tRNA

Question 32

Translation step 1

Answer 32

charged tRNA carrying the next amino acid to be added to the polypeptide chain binds to the vacant A site

Question 33

Translation step 2

Answer 33

polypeptide chain on site P attaches to A site to add to chain

Question 34

Translation step 3

Answer 34

large subunit moves over the chain to open up the A site

Question 35

Translation step 4

Answer 35

tRNA ejection from E site to allow process to repeat

Question 36

Initiation of protein synthesis in eukaryotes requires...

Answer 36

translation initiation factors and a special initiator tRNA

Question 37

Translation halts at a

Answer 37

Stop codon *the binding of release factor to an A site bearing a stop codon terminates translation of an mRNA molecule.

Question 38

A single prokaryotic mRNA molecule can encode

Answer 38

several different proteins.

Question 39

Bacterial translation

Answer 39

can start at a variety of different locations allowing multiple different proteins from one RNA

Question 40

small and large ribosomal RNA only form together after

Answer 40

small ribosomal subunit has bound to mRNA

Question 41

tRNA have ________that is antiparallel to the codon

Answer 41

anticodons

Question 42

Translation is initiated by a complex of:

Answer 42

* translation initiation factors * initiator tRNA * small subunit of the ribosome

Question 43

Proteasome...

Answer 43

degrades proteins marked by ubiquitin

Question 44

many proteins require _____ after translation

Answer 44

modifications to become functional

Question 45

Protein modification

Answer 45

non covalent folding Covalent - phosphorylation - glycosylation

Question 46

The final concentration of each protein depends on...

Answer 46

the rate of each step depicted.

Question 47

Regulation of gene expression allows for the

Answer 47

differences between cells with the same genome.

Question 48

Many transcription regulators bind to DNA as _______, rather than monomers

Answer 48

dimers

Question 49

Activator protein

Answer 49

binds to enhancer and attracts poly II

Question 50

Repressor

Answer 50

decrease transcription by blocking assembly or preventing RNA pol moving forward.

Question 51

Combinatorial Control

Answer 51

Combinations of a few transcription regulators can generate many cell types during development.