BIOMG 3300: Principles of Biochemistry - Unit 7

Question 1

Describe the function of each of the following (p. 281):

a. rRNA
b. mRNA
c. tRNA

Answer 1

a. Ribosomal RNAs- components of ribosomes, the complexes that carry out the synthesis of proteins.
b. Messenger RNAs-intermediates, carrying genetic information from one or a few genes to a ribosome, where the corresponding proteins can be synthesized.
c. Transfer RNAs- adapter molecules that faithfully translate the information in mRNS into a specific sequence of amino acids.

Question 2

Compare DNA and RNA with respect to:

a. Identity of the pentose ring (ribose vs deoxyribose; Fig. 8-1, p. 281).
b. Which nitrogen bases (C,A,G,U, and/or T) are found in DNA? In RNA? (Fig. 8-2, p. 282)

Answer 2

a. In DNA the atom at the 2’ carbon is a hydrogen. In RNA there is a hydroxyl group attached to the 2’ carbon.
b. In DNA: cytosine, thymine, adenine, and guanine are the nitrogen bases present. In RNA, Cytosine, Adenine, Guanine and URACIL are the nitrogen bases present.

Question 3

Distinguish between monocistronic and polycistronic mRNA (p. 294). Is eukaryotic mRNA monocistronic or polycistronic? What about prokaryotic mRNA?

Answer 3

• Monocistronic mRNA encodes for one polypeptide or gene. Polycistronic encodes for multiple polypeptides or genes.
• Eukaryotic mRNA is mostly monocistronic
• Prokaryotic mRNA can be either monocistronic or polycistronic

Question 4

Secondary structure (pp. 294 - 296)
 a. Is DNA usually double or single stranded?
 b. Is RNA usually double or single stranded (p. 294)?
 Does single stranded RNA have any double stranded regions?
 Using Fig. 8-23 (p. 295) to illustrate your answer, point out a hairpin loop (sometimes called a stem and loop).
 Within double-stranded regions of RNA, define two types of base pairs.
 Note: The molecule shown in Fig. 8-24 (p. 295) has much more secondary structure than would be found in a typical messenger RNA.

Answer 4

a. DNA is usually double stranded.
b. The product of transciption is always single -stranded RNA. The RNA molecule has a high tendancy to fold and pair up amongst its complementary regions.
- bulges usually result when base pairs are unmatched or mismatched.
- hairpin loops form from self complimentary palindromic sequences that fold and pair with themselves.

Question 5

Use Fig. 26-1 (p. 1058) to give an overview of transcription by RNA polymerase.
Distinguish between RNA polymerases and DNA polymerase I in terms of :
a.Substrates
b.Whether a primer is needed
c.The direction of synthesis (3’→5’) or (5’→3’)
d.The nucleophilic attack in bond formation
e.The role of pyrophosphate and pyrophosphatase in the overall reaction
f.The requirement for a template

Answer 5

a. RNA is synthesized via a DNA template, It also requires (Adenosine,cytosine, guanine, uracil)- triphosphate nucleotides.
b. RNA polymerase does not need a primer
c. RNA polymerase syntehsizes RNA in the 5’ ->3’ direction.
d. The 3’ hydroxyl group acts as a nucleophile attacking the alpha phosphate of the incoming triphosphate.
e. The pyrophoshate is stabilized and removed by two mg²⁺ ions.
f. RNA polymerase requires a DNA template strand

Question 6

What is meant by the terms “template” and “non-template” strands (Fig. 26-2, p. 1059).?

Answer 6

There are two complementary strands in DNA. The RNA transcript is synthesized on the template strand and is identical to the complementary DNA strand.

Question 7

Does the same strand of the chromosome always serve as the template strand(Fig. 26-3, p. 1059)?

Answer 7

The coding strand for a particular gene may be located in either strand of a given chromosome.

Question 8

Discuss the quaternary structure of RNA polymerase (Fig.26-4, p. 1060).

Answer 8

A large complex enzyme with five core subunits. Ths shape is that of a crab claw. The pincers are composed of the beta subunits

• Two alpha subunits
• a beta subunit
• a beta prime subunit

A sixth subunit known as the sigma unit binds transiently to the core and directs the enzyme to specific biding sites on the DNA. These six enzymes constitute the RNA polymerase holoenzyme.

Question 9

Does RNA polymerase have 3’ to 5’ exonuclease activity? Why is it unnecessary (p. 1060)?

Answer 9

RNA polymerase lacks a proofreading exonuclease. This is unnecessary because many copies of an RNA are generally produced froma single gene and all copies or RNA are eventually degreaded and replaced. Whereas DNA needs the proof reading mechanism because it is the permenant repository of genetic information.

Question 10

Discuss the general organization of an E. coli promoter. In doing so, consider the following (pp. 1060-1061):
a.What is meant by the term “consensus sequence”? Use Fig. 26-5(p. 1061) to discuss how a consensus sequence is determined? Note that the term “consensus sequence” is a general term. In Fig. 26-5 we are focusing on the consensus found at prokaryotic transcription
initiation sites. There are, however, many different consensus sequences with many different functions.
b.Define the numbering system which is used to identify regions such as
the-10 and-35 regions.
c.What is the significance of the-35 and-10 regions (p. 1060)?
d.What is the relationship between the “strength” of a promoter (i.e. the frequency at which transcription is initiated at a particular promoter) and the degree of match to a consensus sequence (p. 1060)?
e.What is the function of the sigma (σ) subunit (p. 1060)?

Answer 10

RNA polymerae binds to specific sites on the RNA called promoters

a. comparisons of common bacterial promoters have revealed common short sequences centered at -10 and -035. These sequences are rich in Adenosine and Tyrosine
b. Base pairs before the start of RNA synthesis is denoted by negative numbers.. base pairs after the end of RNA synthesis is denoted by positive base pairs.
c. These are two concensus regions that are some of the most commoon class of bacterial promoters These are important interaction sites for the sigma subunit.
d. The more similiar the consensus sequence, the more efficent the promoter. A change in at least one base pair can deduce the strength of the promoter in an order of magnitude.
e. The sigma factor directs the bound polymerase to it’s promoter

Question 11

Use Fig. 26-6 (p. 1063) to discuss the events in transcription initiation.

Answer 11

1. RNA polymerase is directed to the promoter by it’s bound sigma transcription factor.
2. A closed complex is formed between the polymerase and the promoter.
3. More upstream on the DNA, an open complex of unwound DNA begins to form.
4. transcription is then initiated. This causes the complex to move away from the promoter as elongation continues.
5. The sigma subunit then dissociates as elongation proceeds, causing the protein NusA to bind with the associated complex ( in the basence of the sigma sub unit)
6. when transcription is over the NusA dissociates, as does the polymerase from the DNA.

Question 12

Is transcription usually regulated at the level of initiation, elongation, or termination (p. 1061). Is it ever regulated at the other levels?

Answer 12

Regulation can occur at all steps. Although it is usually directed towards the initiation steps. ( Particularly with polymerase binding and trasncription initiation.)

Question 13

Discuss two distinguishing features of the termination signal in the mRNA in rho independent termination of transcription in prokaryotes (Fig. 26-7a, p. 1065).

Answer 13

• Rho independent terminators have two distinguishing features. The first is a transcript with highly complementary features that fold back upon themselves and form a hairpin structure
• The second is a highly conserved string of three A residues in the template strand that are transcribed into u residues near the 3’ end of the hairpin. The hair pin disrupts the A_U base pairs causing the RNA and RNA polymerase to dissociate.

Question 14

Transcription in eukaryotes (pp. 1064-1068)

a. Using Fig. 26-8 (p. 1065), describe the sequences typically found at a eukaryotic promoter. How do these sequences compare to those found at a prokaryotic promoter?
b. What is a transcription factor (p. 1066)?

Answer 14

• Eukaryotic promoters usually have a TATA box ( a consesus sequence) near base pair 30. Eukaryotic Promoters also have an INR sequence near the RNA start site at 1.
• Transcription factors are proteins that help to form the active complex.

Question 15

Use Fig. 26-11 (p. 1070) to discuss the formation of a primary transcript and its processing to an mRNA.

Answer 15

A newly syntesized RNA molecule is called a primary transcript

A 5’ cap is added before the synthesis of the primary transcript is complete. A non coding intron sequence is added to the 3’ end. This is known as a poly (A) tail.

Once the primary transcript has been completed, cleave and polyadenylation processing occurs to create the mature mRNA

Question 16

During the processing of eucaryotic mRNA a “5’ cap” is added (Fig. 26-12,,p. 1071). Describe and list two functions of the cap (p. 1070)

Answer 16

• The 5’ end cap has a special bond that protects the mRNA from the activity of ribonucleases ( enzymes the degrade RNA)
• The 5’ endcap also helps bind the mRNA to the ribosome so it can proceed with translation.

Question 17

Splicing of the primary transcript (pp. 1070-1075):

a. Distinguish between “exons” and “introns” (p. 1070).
b. Do all eukaryotic genes contain introns (p. 1070)? Do most?
c. Does splicing take place at the level of DNA or RNA?

Answer 17

• Introns are noncoding segments fo RNA, while exons are coding sequences of RNA.
• Most eukaryotes contain introns in their sequences for polypeptides
• Most splicing takes place at the level of RNA.

Question 18

d.Discuss the splicing mechanism in mRNA shown in Fig.26-16(p. 1074).

Answer 18

A

1. At the 5’ end of the intron: there is a GU sequence. An A branch and a purimidine rich region adn at the 3’ end there is an AG sequence.
2. The U1 subunit binds to the Gu sequence. The U2 subunit binds to the A sequence. ( This becomes the nucleophile during the splicing sequence.)
3. The U1 and U2 snurps bind, and the remaining (4,5,6) bind to form an inactive spliceosome.
4. An internal arrangment occurs expelling the U1 and U4 subunit.
5. Coordination brings the two spliceosomes units together.
6. The 3’ OH becomes the nucleophile that splices the 5’ phosphodiester bond.

Question 19

1)Point out features at exon-intron junctions that provide signals for correct splicing. What are the consequences of imprecise splicing?
2)What is a spliceosome? What is an snRNP (also called a snurp)? 3)How do individual snRNAs locate the consensus sequences that flank introns (p. 1073)?
4)Point out the location and identity of the nucle
otide that will make the initial attack on the 3’ end of the exon.
5)What is the “lariat structure”?

Answer 19

1. A GU site at the 5’ end. An A branch sequence with a pyrimidine rich region. An AG sequence at the 3’ end. The consequences for incorrect splicing is a completely different product processed in a different manner.
2. A splicesome is a large protein complex that splices introns. An snRNP specialized protein-RNA complexes.
3. snRNA’s usually have a complementary sequences near the splice sites.
4. The Adenosine base neares the 3’ end of the intron
5. The structure is a loop formed by U2 and U6

Question 20

Another modification of eukaryotic mRNA is addition of a 3’ poly (A) tail (p. 1075).

a. What is the function of the poly (A) tail (p. 1075)?
b. Discussthe key steps in the addition of a poly (A) tail (Fig. 26-17,p. 1075).
1) Is the tail encoded in the gene?
2) Name the enzymes involved in making the poly (A) tail.
3) Does this enzyme add residues in a template directed manner like DNA polymerase?

Answer 20

• The poly(A) tail serves as a a binding site for one or more specific proteins. The poly(A) tail and it’s asociated proteins probably help mRNA from enzymatic destruction.
• The RNA is cleaved by endonuclease 10- 30 nucleotides 3’ downstream of the AAUAAA cleavage signal. The polyadnylate polymerase synthesizes a poly(A) of 80-250 bp long.
• The enzyme does not add residues in a template fashion.

Question 21

Clearly distinguish between the processes shown in Fig. 26-18 (p. 1076) and “RNA editing” described briefly in the text (p. 1075) and covered more fully
in Unit 8?

Answer 21

• The figues implies that there is only one pathway for the RNA processing to proceed. In reality, the RNA can be processed in multiple ways.

Question 22

6.Use Fig. 26-19 (p. 1076) to distinguish between poly A site choice and alternative splicing. Discuss the role of each of these mechanisms in generating protein diversity from a single gene. Note that because there are substantially fewer genes in the human genome (~25,000) than the number of different proteins made (>50,000), it is believed that many genes can actually encode more than one protein through these types of mechanisms.

Answer 22

If there are two or more sites for polyadenylation and clevage, then multiple products can be formed from the same transcript.

Question 23

RNA-dependent synthesis of RNA and DNA (pp. 1085 -1098)

a. Use Fig. 26-32 (p. 1086) to discuss how retroviruses (like HIV) infect host cells. What is the role of reverse transcriptase?
b. AZT is a common component in the triple cocktail used to treat HIV positive patients. Discuss why AZT interferes with viral replication (Box 26-2, p. 1089).
c. What is meant by “RNA world”. What evidence is used in support of the hypothesis (pp. 1092-1095)?

Answer 23

• The Reverse transcriptase synthesizes DNA from RNA.
• AZT competitvely inhibits the reverse transcriptase. When it is integrated into the DNA sequence, it does not have a 3’ hydroxyl to continue elongation.
• The RNA world hypothesis theorizes that RNA might have been the precursor to DNA.
  
  • Catalytic RNAs
  • pathways for interconversion from rna to dna

Question 24

What percentage of the human genome is believed to encode protein
(Box 26-4, pp. 1096-1097)? How much of our genome is currently believed to be transcribed into RNA? What is a “TUF”?