2.2 Micro

Question 1

Transcription

Answer 1

the synthesis of an RNA molecule
complementary and antiparallel to one of the two strands of a double-
stranded DNA molecule.

Question 2

three main products of transcription

Answer 2

messenger RNA (mRNA),
transfer RNA (tRNA), and ribosomal RNA (rRNA).

Question 3

RNA polymerase (RNAP)

Answer 3

synthesizes RNA in the 5’→3’ direction

Question 4

Bacterial RNAPs consist of five subunits

Answer 4

β, β’, α, ω, and σ, with α present in two subunits.

Question 5

RnaP core
enzymes structure and function

Answer 5

α2, β, β’,σ, the function is transcription elongation

Question 6

RnaP holoenzyne
enzymes structure and function

Answer 6

α2, β, β’,σ, ω, the function is it recognizes promoters

Question 7

When is transcription initiated?

Answer 7

When the RnaP holoenzyme binds to the DNA promoter region upstream of the transcription start site.

Question 8

What recognizes the promoter anf what reigons?

Answer 8

σ factor, it recognizes the -10 region and the -35 region

Question 9

After the holoenzyme binds….

Answer 9

RNAP opens the DNA helix and
unwinds a short portion of DNA, creating a transcription bubble.

Question 10

after transcription has proceeded…

Answer 10

sigma factor falls off
and the core enzyme continues transcription.

Question 11

RpoD (σ70) is the sigma factor responsible for transcription of most genes, including the major housekeeping
genes for normal growth. Not all genes expressed by RpoD are transcribed with equal efficiency. What features
might determine the efficiency of transcription by RpoD?

Answer 11

depends on how closely the promoter matches the σ⁷⁰ consensus sequences. The more closely it matches, the more efficient the transcription

Question 12

Rho-dependent termination

Answer 12

relies on the binding of a protein (Rho) to regions on RNA. RNA is then threaded through the Rho protein,
pulling Rho towards the RNA polymerase. When Rho contacts RNAP, it causes its dissociation, thus terminating transcription.

Question 13

Rho-independent termination

Answer 13

relies on the formation of a stem-loop structure. The weak U-A binding between the RNA and the DNA template causes the polymerase to pause. NusA protein binds the GC-rich stem loop, pushing RNAP off the RNA, thus terminating transcription.

Question 14

Translation

Answer 14

the synthesis of protein by a ribosome using the genetic information in a mRNA as a template.

Question 15

Ribosomes

Answer 15

catalyze the linkage of amino acids during
translation, using mRNA as the code and charged tRNAs as the source of amino acids

Question 16

The small subunit (30S)

Answer 16

contains 21 rProteins assembled around one 16S rRNA molecule
the latter of which forms the channel for mRNA and the binding sites for tRNA

Question 17

The large subunit (50S)

Answer 17

contains 33 rProteins assembled around two rRNA molecules (5S and
23S

Question 18

23S RNA

Answer 18

mediates the peptidyltransferase activity of the ribosome—formation of the amino
acid linkages (peptide bonds).

Question 19

Aminoacyl-tRNA synthetase

Answer 19

facilitates tRNA charging

Question 20

TRNA charging steps

Answer 20

Amino acids are first activated by a reaction with ATP, forming an aminoacyl-AMP molecule.
Amino acids are then transferred to the hydroxyl residue of the terminal adenine of the tRNA by aminoacyl-tRNA synthetase, releasing AMP.

Question 21

What does translation initiation require?

Answer 21

Initiation factors

Question 22

Translation initiation steps

Answer 22

Initiation factors bind the 30S subunit, causing dissociation of the two ribosome subunits.
Initiation factors then facilitate the binding of the mRNA to the 30S subunit.
Next, initiation factors escort the initiator tRNA to the start codon in the P site.
The first amino acid incorporated is a modified form of methionine with an added formyl group, termed N-formyl methionine.
Finally, initiation factors dissociate, allowing the large subunit to bind, finalizing the formation of the
initiation complex.

Question 23

Translation elongation requires the activity of small proteins called

Answer 23

elongation factors

Question 24

Translation elongation steps

Answer 24

• Elongation factors bind tRNA molecules and
  carry them to the A site. They are subsequently released.
• Peptidyltransferase activity catalyzes a peptide bond between the amino acid in the P site and the amino acid in the A site, with the peptide being transferred from the tRNA in the P site to the tRNA in the A site.
• Elongation factors translocate the ribosome down one codon, shifting the empty tRNA to the E site and the peptide-holding tRNA to the P site
• The empty tRNA exits the E site and another round of amino acid addition can begin.

Question 25

Translation termination requires the activity of small proteins called

Answer 25

release factors

Question 26

Translation termination steps

Answer 26

- When the mRNA stop codon enters the A site, release factors bind, activating peptidyltransferase to cut the final bond and release the polypeptide. - Release factors then eject the final tRNA from the ribosome.

Question 27

You experimentally change the DNA sequence directly upstream of a start codon of an operon in E. coli to investigate the function of this region of DNA. Analysis reveals that after the change, the same amount of mRNA is made from the operon, but there are very few proteins made from the operon. What is the most likely function of the DNA sequence that you changed? a. Ribosome-binding site b. Promoter c. Termination sequence d. Transcriptional regulation

Answer 27

Ribosome-binding site because the problem is with translation meaning there is a problem with the ribosome binding site

Question 28

Some essential genes and DNA sequences in cells do NOT encode for proteins but are still essential for cellular growth and replication. Give two examples of a gene or sequence for which this is TRUE and explain why it is essential for growth or replication.

Answer 28

rRNA - it doesn't code for proteins, but it does code for ribosomal RNA, and without it the cell cannot make functional ribosomes which means no protein synthesis and therefore no growth or replication Oric sequence - Without a functional oriC, DNA replication cannot initiate, so the cell cannot duplicate its genome → no cell division or growth.

Question 29

Proteins destined for secretion are differentiated by

Answer 29

specific signal sequences recognized by cellular transport systems

Question 30

Membrane proteins (Signal sequence recognized by, Transmembrane transporter, Destination, Timing of transport relative to translation)

Answer 30

SRP, SEC YEG, membrane, during translation

Question 31

unfolded proteins (Signal sequence recognized by, Transmembrane transporter, Destination, Timing of transport relative to translation)

Answer 31

SecA, SecYEG, Periplasm, after translation

Question 32

pre-folded proteins (Signal sequence recognized by, Transmembrane transporter, Destination, Timing of transport relative to translation)

Answer 32

Tat BC, Tat A, periplasm, during translation

Question 33

In Gram-negative bacteria whats required when transporting proteins across the outer membrane

Answer 33

specialized secretion sytems (I, II, and V)

Question 34

Type I secretion systems

Answer 34

one-step systems that contain a channel spanning both inner and outer membranes.

Question 35

Type II and Type V systems

Answer 35

two-step systems that rely on Sec or Tat systems to move proteins from the cytoplasm to the periplasm, followed by outer membrane proteins to transport from the periplasm to the extracellular space.

Question 36

Type III, IV, and VI secretion systems

Answer 36

one-step secretion systems that transport proteins or DNA directly across host cell membranes (e.g., during infection or during gene transfer).

Question 37

Type IV secretions systems

Answer 37

can secrete proteins and DNA into the extracellular space in addition to across host cell membranes. These are the most widespread secretion systems.

Question 38

Type VI secretion systems

Answer 38

use a needle-like projection to propel the secreted protein across the bacterial membrane and across the host membrane.

Question 39

You are studying a protein in Salmonella typhimurium that you believe is a toxin. Whenever you attempt to purify the protein from lysed cell cultures, you get two forms of the protein. One form is smaller than the other and is missing 15 amino acids from the N-terminus compared to the larger form. This leads you to hypothesize that There are two termination sites in the mRNA The protein requires specialized machinery to fold properly The protein is secreted and folds outside the cell

Answer 39

The protein is secreted and folds outside the cell