Ch 8

Question 1

Describe how DNA serves as genetic info

Answer 1

the structure of DNA helps explain two primary features of biological info storage.

-first, the linear sequence of bases provides the actual info

-genetic info is encoded by the sequence of bases along the strand of DNA, in much the same way as our written language uses linear sequence of letters to form words and sequences

Question 2

Describe and recognize the process of DNA replication (aka DNA synthesis), including the function of the necessary enzymes and directionality of replication. Be able to repeat the replication fork

Answer 2

DNA synthesis begins at the origin of replication (modeled
as oriC) in prokaryotes

Formation of a replication fork must occur: zone of
unwound DNA where DNA polymerization is occurring

Question 3

steps in DNA synthesis (DNA replication)

1. DNA topoisomerases like DNA gyrase

Answer 3

remove the supercoils in the DNA to create linear DNA

Question 4

steps in DNA synthesis (DNA replication)

1. DNA topoisomerases like DNA gyrase

Answer 4

remove the supercoils in the DNA to create linear DNA

Question 5

steps in DNA synthesis (DNA replication)

2. DNA is nicked: DNaA proteins bind to the

Answer 5

origin and cleave the hydrogen bonds between complementary strands

Question 6

steps in DNA synthesis (DNA replication)

3. Next, DNA helicase acts to unwind

Answer 6

the DNA for a short distance and expose it as 2 single-stranded regions

Question 7

steps in DNA synthesis (DNA replication)

4. next, single-stranded binding proteins (SSBs) bind to these exposed

Answer 7

regions and prevent them from re-annealing to one another

Question 8

steps in DNA synthesis (DNA replication)

5. an RNA primer is synthesized against the template

Answer 8

ssDNA by primase

Question 9

steps in DNA synthesis (DNA replication)

6. Extension of DNA or polymerization (adding 3’ end) is performed by DNA polymerase III

Answer 9

occurs continuously on the leading strand

occurs discontinuously on the lagging strand creating okazaki fragments
new primers have to be added as
the fork ‘moves’ and exposes new
DNA ahead of the primer

Question 10

steps in DNA synthesis (DNA replication)

7. Once DNA synthesis is complete along the template the RNA primers must be

Answer 10

removed and replaced by DNA nucleotides
the RNA primers are removed by
DNA polymerase I proofreading
after the synthesis step

Question 11

steps in DNA synthesis (DNA replication)

8. Okazaki fragments are sealed together by DNA ligase to form

Answer 11

a continuous piece of DNA (no RNA nucleotides)

Question 12

semi-conservative DNA replication

Answer 12

each chromosome now has 1 single strand of parental DNA (template) and one newly synthesized strand of DNA (daughter)

Question 13

major enzymes in DNA synthesis

helicase

Answer 13

unzipping the DNA helix

Question 14

major enzymes in DNA synthesis

gyrase

Answer 14

helping to untangle the DNA supercoils

Question 15

major enzymes in DNA synthesis

primase

Answer 15

synthesizing an RNA primer

Question 16

major enzymes in DNA synthesis

DNA polymerase III

Answer 16

adding bases to the new DNA chain; proofreading the chain for mistakes

Question 17

major enzymes in DNA synthesis

DNA polymerase I

Answer 17

removing primer, closing gaps, repairing mismatches

Question 18

major enzymes in DNA synthesis

ligase

Answer 18

final bonding of nicks in DNA during synthesis and repair

Question 19

Describe and recognize the differences between DNA and RNA.

Answer 19

DNA
- deoxyribose sugar
- double stranded
-bases: Adenine-Thymine Cytosine-Guanine

RNA
-ribose sugar
-single stranded
-bases: Adenine-Uracil Cytosine-Guanine

Question 20

Transcription Steps

1. transcription begins when RNA polymerase bind to the

Answer 20

promoter sequence

Question 21

Transcription Steps

2. transcription proceeds in the

Answer 21

5’->3’ direction

Question 22

Transcription Steps

3. transcription stops when it reaches the DNA-based

Answer 22

terminator sequence

Question 23

Transcription Steps

4. the mRNA transcript looks just like the coding strand (U’s for T’s) because

Answer 23

RNA polymerase adds nucleotides complementary to the template strand (5’ to 3’)

Question 24

What enzyme performs transcription?

Answer 24

RNA polymerase

Question 25

**Steps in Translation: Initiation** 30S and 50S ribosomal subunits assemble with

Answer 25

mRNA and Met-tRNA

Question 26

**Steps in Translation: Initiation** 1. process is aided by several proteins called

Answer 26

initiation factors (IF's)

Question 27

**Steps in Translation: Initiation** 2. begins when the AUG start codon is situated in the

Answer 27

P site (peptidyl site) of the ribosome

Question 28

**Steps in Translation: Initiation** 3. allows the anticodon portion of Met-tRNA to bind to the

Answer 28

AUG of the mRNA

Question 29

**Steps in Translation: Elongation** amino acids are brought into the ribosome to

Answer 29

elongate the polypeptide, then the ribosome is translocated to a new codon

Question 30

**Steps in Translation: Elongation** 1. A (aka receptor) site is filled:

Answer 30

Aminoacyl tRNAs (tRNA + amino acid) are brought to the A site of the ribosome (specific amino acid selected by complementary tRNA anticodon- mRNA codon binding) this is aided by protein elongation factors (EFs)

Question 31

**Steps in Translation: Elongation** 2. Transpeptidation

Answer 31

the amino acids in the P site join the new amino acid in the A site via formation of a peptide bond between aa's

Question 32

**Steps in Translation: Elongation** 3. Translocation

Answer 32

movement of the ribosome along the mRNA occurs to move a new codon into the A site

Question 33

**Steps in Translation: Elongation** 3. Translocation how does it work?

Answer 33

A site --> P site P site --> E site (aka exit) A site will be emptied so a new aminoacyl tRNA can enter it uncharged tRNA in the E site dissociates from the ribosome completely

Question 34

**Steps in Translation: Termination** occurs when a stop codon enters the

Answer 34

ribosome's A site aided by protein release factors (RF's) which recognize stop codon

Question 35

**Steps in Translation: Termination** 1. completed polypeptide is cleaved from tRNA and ribosome subunits

Answer 35

dissociate

Question 36

**Steps in Translation: Termination** 2. subunits are free to form new

Answer 36

initation complexes and repeat the entire process on a new mRNA strand

Question 37

coupled transcription/translation,

Answer 37

Prokaryotes are much more efficient at protein expression than eukaryotes * Since there is no membrane to separate the DNA from the cytoplasm, translation of mRNA can start as soon as a transcript is produced

Question 38

transpeptidation

Answer 38

the amino acids in the P site join the new amino acid in the A site via formation of a peptide bond between aa’s

Question 39

translocation

Answer 39

movement of the ribosome along the mRNA occurs to move a new codon into the A site

Question 40

mutation:

Answer 40

change in genetic material- within the nucleotide sequence

Question 41

point mutation:

Answer 41

a single base (nucleotide) change only affects 1 codon

Question 42

missense mutation

Answer 42

results in a different amino

Question 43

silent (aka neutral) mutation

Answer 43

results in the same amino acid code still being encoded

Question 44

nonsense mutation

Answer 44

results in the conversion of an affected codon from endoding an amino acid to becoming a stop or nonsense codon

Question 45

frameshift mutation

Answer 45

results in multiple changes

Question 46

transformation

Answer 46

Naked (free) DNA/plasmids in the environment can be taken up by a cell * For successful HGT & recombination: this DNA must recombine into chromosome by homologous recombination * Exception: a whole plasmid can be taken up and remain free in the cytoplasm! * Non-homologous genes will be degraded (= unsuccessful HGT/recombination)

Question 47

transduction

Answer 47

HGT mediated by a bacteriophage (bacterial virus) carrying DNA into a host as it ‘infects’ the host * A incorrectly packaged bacteriophage carries bacterial chromosomal DNA into the cell it infects * Called a transducing phage * Cell can gain new genes via homologous recombination of the introduced DNA (if HGT is successful) * If there is no homologous genes in the new host, the transferred DNA could also be degraded (this would be unsuccessful HGT)

Question 48

transposons

Answer 48

Segments of DNA that can move from one region of DNA to another = ‘jumping genes’ * Contain a gene for the transposase, enzyme that helps it move (for cutting and resealing DNA) * Insertion sequences only have the transposase gene * Composite (complex) transposons carry other genes plus transposase (which is often sandwiched by two insertion sequence

Question 49

conjugation

Answer 49

Cell-to-cell transfer of a plasmid. The original cell retains the plasmid and one strand moves into the recipient cell (where the other strand is replicated) * Also called mating but its not sexual reproduction! Requirements for conjugation: * Cell-to-cell contact between donor & recipient cells via a pilus made by the donor cell * tra genes - these encode the pilus and other apparatus necessary for conjugation * oriT (origin of transfer): this is the site where rolling-circle replication begins to make a copy of the plasmid DNA that is exchanged with the neighboring cell (in contact via pilus)