transcription, translation, replication

Question 1

transcription: central dogma

Answer 1

how we get from DNA to the expression and creation of proteins. describes the flow of information.

Question 2

transcription:

Answer 2

transcription takes place in the nucleus. translation happens in the ribosomes.
transcription- synthesis of RNA from a DNA template. RNA is complementary to DNA template. catalysed by RNA polymerase. occurs in the nucleus and the products is mRNA.

Question 3

RNA polymerase:

Answer 3

catalyses the initiation and elongation of RNA chains. 5’->3’ polymerase activity.
requires- DNA template (double or single stranded), all 4 ribonucleoside triphosphates (ATP, GTP, CTP, UTP), divalent metal ion (mg2+ or Mn2+).

Question 4

stages of transcription:

Answer 4

1st transcribed nucleotide is usually a purine.
3 phases- initiation, elongation, termination.
initiation occurs at the promoter.
RNA polymerase and the sigma factor bind at the promoter.
terminator codes for a sequence that forms hairpin structure followed by a strong of uridines.

Question 5

termination of transcription:

Answer 5

DNA template has start and stop signals. termination involves several processes- transcription stops, RNA-DNA hybrid dissociates, melted region of DNA rewinds. RNA polymerase releases DNA.

Question 6

types of RNA:

Answer 6

all types of RNA transcribed from a DNA template in manner described.
main 3 types of RNA- messenger RNA (mRNA), transfer RNA (tRNA), ribosomal RNA (rRNA).

Question 7

messenger RNA (mRNA):

Answer 7

5% of cellular RNA. no specific secondary structure (can form hairpin loops that regulate its lifespan). sequence of bases containing the information for the sequence of amino acids in the protein to be synthesised. mRNA is the template for protein synthesis. variable size and sequence. transcribed from protein coding genes.

Question 8

the TATA box promoter- the main promoter in eukaryotic cells:

Answer 8

this sequence is normally located 25bp upstream of the transcription start site. it is AT rich and binds a number of proteins including RNA polymerase II. the proteins bind in sequence and do so to stabilise other proteins

Question 9

ribosomal RNA (rRNA):

Answer 9

80% of cellular RNA. several forms- prokaryotes have 23S, 16S and 5S rRNA. called S because of their sedimentation behaviour. one molecule of each rRNA species is present in each ribosome. major component of ribosomes (the structures on which proteins are synthesised.

Question 10

transfer RNA (tRNA):

Answer 10

25% of cellular RNA. carries and delivers amino acids in an activated for the ribosomes for peptide bond formation during protein synthesis. at least one tRNA for each of the 20 amino acids. tRNA contain an amino acids attachment site and a template recognition site called an anticodon.

Question 11

translation:

Answer 11

synthesis of proteins. the sequence of bases in mRNA specifies the sequence of amino acids in the protein product.
takes place on ribosomes- large protein rRNA complexes, 2 subunits of unequal size, subunits in prokaryotes and eukaryotes differ.

Question 12

binding sites on ribosome:

Answer 12

3 binding sites-
P (peptidly-tRNA)
A (aminoacyl-tRNA)
E (exit)

Question 13

the genetic code:

Answer 13

an amino acid is coded for by a group of 3 bases called a codon. there is no overlap of codons. the code is read continuously from a fixed starting point. the code is degenerate- 1 amino acid may be coded for by several codons. the code is unambiguous- a codon only codes for 1 amino acid. the code has start and stop signals. the code is almost universal.

Question 14

tRNA:

Answer 14

delivers amino acids to the ribosome.
each tRNA delivers a single specific amino acid.
anticodon of t RNA recognises and base pairs with complementary codon of mRNA.

Question 15

wobble pairing of tRNA with mRNA:

Answer 15

there are 61 amino acid coding codons but only 40 tRNA molecules. some tRNA recognise more than 1 codon (same amino acid). 5’ end of tRNA can have non standard base pairing- wobble pairing.

Question 16

initiation of translation:

Answer 16

AUG is initiation sequence. methionine in eukaryotes, formylmethionine (fMet) in prokaryotes. in prokaryotes, GUG can also be initiation sequence. translation does not start immediately at the 5’ end of the mRNA. start is nearly always 25 nucleotides away from 5’ end. reading frame is established after the initiator AUG has been located.

Question 17

initiation sequence:

Answer 17

in prokaryotes- needs purine rich sequence on the 5’ side of the initiator sequence called shine dalgarno sequence.
in eukaryotes- needs kozak sequence followed by AUG (AUG closest to 5’ end).
initiation sequence helps recruit the ribosome to the mRNA to initiate protein synthesis by aligning it with the start codon.

Question 18

termination of translation:

Answer 18

stop codes (UAA, UAG, UGA) designate chain termination. these codons are not read by tRNA but are read by specific proteins called release factors. binding of the release factor to the ribosome releases the newly synthesised protein.

Question 19

translation occurs using multiple ribosomes:

Answer 19

mRNA is translated in the 5’ to 3’ direction. proteins are synthesised from the amino to the carboxyl end. many ribosomes can translate a single mRNA at the same time.

Question 20

DNA replication:

Answer 20

DNA serves as a template for its own synthesis. DNA is semi conservative.

Question 21

enzymes involved:

Answer 21

in 1958 kornberg and colleagues first isolated DNA polymerases from E coli. Promote formation of phosphodiester bonds joining units of DNA backbone. The nucleotide added is the one able to form a base pair with the next nucleotide of the template. There are several DNA polymerases. Can remove mismatched nucleotides in DNA (repair).

Question 22

DNA polymerase I (pol I):

Answer 22

5’->3’ polymerase activity- addition of deoxyribonucleotides to the 3’ end of a growing DNA chain. DNA synthesis requires – the 4 deoxynucleoside 5’ triphosphates (dATP, dGTP, dCTP, TTP), Mg2+, a DNA template, a primer strand with a free 3’-OH must already be bound to the template stand.

Question 23

primer removal by DNA polymerase I:

Answer 23

during DNA repair. 5’->3’ exonuclease activity: hydrolysis of terminal phosphodiester bond at 5’ end of DNA chain.

Question 24

other enzymes:

Answer 24

DNA polymerase 3- 5’->3’ polymerase activity. 3’->5’ exonuclease activity.
DNA ligase- forms a phosphodiester bond between 3’=OH at the end of one DNA chain and 5’-PO4 at start of another DNA chain. Joins 2 DNA chains together.
DNA primase- DNA synthesises the primer- a short stretch of RNA.
Helicase and gyrase- unwinds the DNA by breaking hydrogen bonds between strands.

Question 25

DNA replication:

Answer 25

leading strand- helicase unwinds DNA. DNA Polymerase makes new DNA in 5' to 3' direction on leading strand. lagging strand- DNA polymerase cant synthesise in 3' to 5' direction. synthesis is in short segments called okazaki fragments.

Question 26

topoisomerases:

Answer 26

during unwinding, the DNA twists build up. Build up of twist would form a resistance that would eventually halt the progress of the replication fork. DNA topoisomerases are enzymes that solve these physical problems in the coiling of DNA. Topoisomerase 1 cuts a single backbone, on the DNA, enabling the strands to swivel around each other to remove the build up of twists. Topoisomerase 2 cuts both backbones, enabling 1 double stranded DNA to pass through another, thereby removing knots and entanglements that can form within and between DNA molecules.

Question 27

model of replication:

Answer 27

replication starts at specific regions called the origins of replication. proteins are recruited to this site forming bubble. helicase and gyrase start unwinding double stranded DNA.

Question 28

action of DNA primase:

Answer 28

DNA primase synthesises RNA primer on both leading and lagging strands.

Question 29

action of DNA polymerase III:

Answer 29

helicase moves along parental DNA unwinding it. DNA polymerase 3 catalyses synthesis of new DNA in 5’->’ direction using 3’-OH groups on RNA primers. leading strand is synthesised continuously by DNA polymerase 3. Lagging strand, primase repeatedly synthesises RNA primers which are then extended by DNA polymerase 3.

Question 30

Joining okazaki fragments:

Answer 30

requires DNA polymerase 1 and DNA ligase. DNA polymerase 1- uses its 5’ to 3’ exonuclease activity to remove the RNA primers. 5’ to 3’ polymerase activity to fill the gaps left by removal of the RNA. Finally, DNA ligase joins the fragments.

Question 31

telomerase- the enzyme:

Answer 31

replication of linear chromosome ends. Composition: RNA and protein. Protein- reverse transcriptase activity- copies RNA sequence. Binds to single stranded region (lagging strand) of linear chromosome and extends the ends (prevents chromosome shortening).