LECTURE 5 - Genetic Code

Question 1

codon

Answer 1

3 bases which code for an amino acid
quoted from 5’ to 3’ from the mRNA
same sequence as non template DNA but is RNA so contains U instead of T

Question 2

start codons

Answer 2

AUG codes for Met which starts the translation

Question 3

stop codons

Answer 3

UAA, UAG UGA

Question 4

point mutation

Answer 4

when one DNA base pair is changed

Question 5

silent mutation

Answer 5

a change in single base pair of dna/in protein coding region that has no effect on protein sequence
type of point mutation

Question 6

nonsense

Answer 6

a single change in DNA code that produces a stop codon so the prematurly terminates the protein syntheisis
type of point mutation

Question 7

insertion

Answer 7

addition of one or more nucleotide base pairs into the DNA sequence

Question 8

deletion

Answer 8

a piece of DNA removed from sequencem

Question 9

missense

Answer 9

a single AMINO ACID is changed

Question 10

frameshift

Answer 10

Insertion or deletion mutation results in a change to a gene’s reading frame`

Question 11

duplication

Answer 11

Incorrect copying leads to repeated sequences

Question 12

central dogma of molecular biology

Answer 12

DNA replicates
DNA to RNA : transcription
RNA to protein : translation

Question 13

copying DNA/formationof phosphodiester bonds

Answer 13

involves template strand from 5’ to 3’
nucleotides triphosphates as substrates,
bind to backbone and then becomes a nucleotides monophosphates
this produces pyrophosphate PPi (P-P and provides energy)
forms a high energy phosphodiester bond

uses DNA polymerase (enzyme)
need a primer to start

Question 14

dna polymerase

Answer 14

make a DNA copy from a DNA template
need a primer to start
Uses deoxynucleotide triphosphates (dNTPs:
dATP [adenine], dGTP [guanine], dTTP [thymine], dCTP [cytosine]) as substrate

Question 15

Topoisomerase/gyrase

Answer 15

enzyme, that relieves supercoiling

causes cuts str
ands, unwinds, and then form together against supercoiling!

Question 16

ORI

Answer 16

where DNA replication starts
its AT rich (2 hydrogen bonds compared to CG which has 3) so easier to break
DNA. binding proteins open up the sites
DNA helicase unwinds part of the DNA
DNA topoisomerase/gyrase stops
supercoiling
Forms replication forks

Single-stranded binding proteins coat single stranded DNA (ssDNA) to keep strands apart/stop small segments of base pairing (hairpins) on the same strang!/protect DNA and not hinder DNA replication= prevent reannealing

RNA primer 5’ to 3’
DNA polymerase III

Question 17

lagging strand

Answer 17

helicase unwinds
Primase makes RNA primer
DNA polymerase III makes okazaki fragments
A different DNA polymerase I removes the RNA primers and fills them in with DNA
And DNA ligase creates a phosphodiester bond between the adjacent DNA fragments joins the DNA fragments

Question 18

termination def

Answer 18

opposite of ORI

Question 19

Joining the ends in circular chromosomal DNA

Answer 19

1) DNA Polymerase III reaches the RNA primer
2) DNA Polymerase I removes the RNA primer and replaces it with DNA
3) DNA ligase joins the ends of DNA
similar to lagging strangd

Question 19

RNA transcription

Answer 19

uses RNA polymerase

Question 19

RNA Polymerases

Answer 19

• Make an RNA copy from a DNA template
• DON’T NEED A PRIMER to start (e.g. primase)
• Use ribonucleotide triphosphates (NTPs: ATP, GTP, CTP, UTP) as substrate
• Limited proofreading, no 3’ to 5’ exonuclease activity (make more mistakes)

Question 20

promoter

Answer 20

RNA polymerase binds to a region of DNA known as the PROMOTER, which sits just
upstream (past to the 5’ end) and starts transcribing downstream from that region.

Question 21

terminator

Answer 21

Transcription stops at the terminator

Question 22

initiationof transcription

Answer 22

1) RNA polymerase binds to the TF and the TF binds to the promoter region
2) Once RNA polymerases has binded to ssDNA, TF leaves

Question 23

Transcription elongation

Answer 23

the two stands of DNA reanneal
as the RNA polymerase passes along, it forms this transcription bubble
5’ to 3’

Question 24

Transcription termination (I)

Answer 24

G/C rich region can form a
ds (double stranded) hairpin structure; causes
transcription to pause and the
RNA polymerase to dissociate
and the RNA to be released

Transcription uses a signal encoded in
the transcribed RNA to stop – a G/C
rich sequence, often followed by an
A/T rich sequence

Question 25

Transcription termination (II)

Answer 25

a) RNA polymerase
dissociates from
the DNA template

b) OR a Rho protein binds, and uses
helicase activity to travel up to and
dissociate the DNA/RNA hybrid complex
(physically pull or force apart the
protein nucleic acid complex)

The RNA
dissociates from
the DNA
template

Question 26

Gene expression regulation

Answer 26

Promotor strength
- DNA sequence optimised for strong or
weak factor (sigma factor in prokaryotic and transcription factor in eukaryotic )/RNA pol binding
- Strong binding = more RNA copies made
- Weak binding = fewer RNA copies made
* Repressors : obstructs RNA polymerase from binding to the promoter and initiating transcription
* Accelerators/Activators: Activators help facilitate the binding of RNA polymerase to the promoter and enhance transcription

Question 27

Transcription Regulation: Repression

Answer 27

A protein repressor binds.
This blocks the binding of the transcription factor/RNApol complex (RNApol is just RNA polymerase)

No RNA polymerase
binding ->
no transcription ->
no gene expression

Question 28

Transcription Regulation: Accelerators

Answer 28

A Transcriptional Activator, a protein, binds at a specific DNA sequence and alters the structure of the promoter so the transcription factor can now bind more frequently

Question 29

Repressors and activators regulation

Answer 29

A ) Trigger change to activate activator

R ) Binding relieves repression allowing sigma factor/RNA pol to bind

Question 30

Semi-conservative replication

Answer 30

each newly synthesized DNA molecule consists of one original (parental) strand and one newly synthesized (daughter) strand.

Question 31

main challenges in copying DNA from DNA and how are they overcome?

Answer 31

1. Unwinding the DNA Double Helix:
   Strategy: Enzymes called helicases are responsible for unwinding the DNA. They use energy from ATP hydrolysis to disrupt hydrogen bonds between base pairs, separating the two DNA strands.
2. Overcoming Supercoiling:
   Strategy: Topoisomerase enzymes, like DNA gyrase in bacteria, help relieve supercoiling by introducing negative supercoils. This prevents excessive torsional strain on the DNA.
3. Initiating DNA Synthesis:
   Strategy: Primase synthesizes a short RNA primer complementary to the template DNA strand. This primer provides the 3’-OH group needed for DNA polymerases to start replication.
4. Coordinating Leading and Lagging Strand Synthesis:
   Strategy: The leading strand is synthesized continuously in the 5’ to 3’ direction. The lagging strand is synthesized discontinuously in Okazaki fragments, and DNA ligase joins these fragments into a continuous strand.
5. Proofreading and Repairing Errors:
   Strategy: DNA polymerases have 3’ to 5’ exonuclease activity, which allows them to proofread and remove incorrectly incorporated nucleotides.
6. Telomere Replication:
   Strategy: Telomerase is an enzyme that adds repetitive DNA sequences to the ends of chromosomes (telomeres) to prevent the loss of essential genetic material during replication.

Question 32

Describe the general functions of proteins that are required for DNA-
Replication

Answer 32

DNA polymerase: responsible for synthesizing new DNA strands during replication, adds nucleotides to the growing DNA strand based on the complementary base-pairing rules and also has proofreading activity to correct errors in replication.
Gyrase: a type of topoisomerase enzyme that helps relieve supercoiling and torsional strain in the DNA molecule by introducing negative supercoils. This is especially important in prokaryotic DNA replication.
Helicase: unwind the double-stranded DNA by breaking the hydrogen bonds between the base pairs, creating two single strands for replication to occur.
Ligase: joining or ligating the Okazaki fragments on the lagging strand during DNA replication. It forms phosphodiester bonds between adjacent nucleotides to create a continuous strand.
Primase: synthesizes a short RNA primer complementary to the DNA template. This primer provides the 3’-OH group necessary for DNA polymerase to initiate replication.

Question 33

challenges of transcription

Answer 33

1. Only small sections of the genome need to be transcribed:
   Strategy: promoters and enhancers, control which sections of the genome are transcribed. Regulatory proteins bind to these elements, guiding RNA polymerase to specific genes.
2. These sections often have to be copied thousands of times:
   Strategy: regulated by the availability of RNA polymerase and the binding of transcription factors. Genes that need to be transcribed frequently have more accessible promoter regions.
3. Some sections are rarely copied in one cell but copied many times in another cell:

Strategy: Differential gene expression is controlled by cell-specific regulatory proteins and chromatin modifications. Different cells have unique transcription profiles, and these variations are achieved through cell-specific regulators. (idkkkk)

Question 34

initation of transcription

Answer 34

RNA polymerase and transcription factors. RNA polymerase x transcription factor attaches to the promoter region. RNA polymerase binds to the promoter region to initiate transcription and TF leaves

Question 35

elongation of transcription

Answer 35

RNA Polymerase travels along the DNA template, following the sequence and building an RNA molecule.

Starting from the initiation site. it unwinds a small section of the DNA double helix, exposing the bases. The RNA polymerase adds complementary ribonucleotides (A, U, C, or G) to the growing RNA chain.
This process is fast and continuous.
The DNA double helix reanneals (joins back together again/ reform) behind the polymerase.

Question 36

Transcription termination

Answer 36

SIGNAL TO STOP : a G/C
rich sequence, often followed by an
A/T rich sequence

G/C rich region can form a
ds hairpin structure; causes
transcription to pause and the
RNA polymerase to dissociate
and the RNA to be released
RNA polymerase and newly formed RNA strand leaves the DNA template

Question 37

proteins that are required for RNA
transcription

Answer 37

RNA Polymerase: reading the DNA template and synthesizing an RNA molecule.

Promoter Recognition Proteins: These proteins recognize specific DNA sequences called promoters, and guide the RNA polymerase to the right place on the DNA strand to begin transcription.

Transcription Factors: can enhance or inhibit the activity of RNA polymerase, depending on the cellular conditions and signals. They ensure that the right genes are transcribed at the right time.

Helicase: unwind the DNA double helix ahead of the RNA polymerase as it moves along the template strand. This unwinding allows the RNA polymerase to access the DNA and create the RNA transcript.

Sigma Factors (in Bacteria): In prokaryotic organisms like bacteria, sigma factors are subunits of RNA polymerase. They play a crucial role in promoter recognition and transcription initiation. Different sigma factors guide the RNA polymerase to different sets of genes, allowing bacteria to respond to changing environmental conditions.

Question 38

transcription regulation

Answer 38

repression : a protein repressor binds to the promoter region so the RNA polymerase complex (RNA polymerase x TF) cannot bind to the DNA sequence –> no transcription

activators : a transcription activator enhances the binding of RNA polymerase to the promoter, promoting transcription.

Question 39

differences between DNA replication and RNA transcroption

Answer 39

Enzyme : DNA vs RNA polymerase
Substrate: existing DNA templates trand vs a specifc region of DNA template strand only
End product : double stranded DNA vs single stranded RNA
Bases : thymine vs uracil
Function : genetic information vs messenger for protein synthesis

FEEBS
function/end product/enzyme/bases/substrate

Question 40

difference between DNA vs RNA polyermase

Answer 40

function : DNA replication vs RNA transcription
end product : DNA double stranded vs RNA single strand
Initiation : uses primer vs uses promoter