C. The Rna Is Read 5' -> 3'; Amino Acids Are Attached to the C-terminus of the Polypetide

ANSWERS to Questions from Office Three

Answers, Affiliate 10. Transcription: RNA polymerases

10.1 The sigma factor (s) causes RNA polymerase to bind to the correct sites on Deoxyribonucleic acid to initiate transcription (i.e. promoters). s destabilizes the complex between core polymerase and not-promoter Deoxyribonucleic acid and decreases the amount of time information technology is bound. It enhances the affinity and increases the amount of time that holoenzyme (a 2 bb's) is leap to promoter, i.e. it facilitates a random search for promoters.

10.ii Statements 2 and 4 are correct.

ten.3 Elongation of transcription by East. coli RNA polymerase proceeds at near l nucleotides per sec. Therefore, the rRNA primary transcript would be synthesized in 6500 nucleotides /50 nucleotides per sec = 130 sec, or slightly over 2 min.

ten.four 0.83 initiations per sec. (50 nt/sec)(three.4 Angstroms/nt) = 170A/sec. 204A/170A sec-1 = 1.ii sec per initiation, or 0.83 initiations per sec.

10.5 a) True

b) Faux

c) True

d) True

10.6 Common features include:

a. All are template directed, synthesizing a sequence complementary to the template.

b. Synthesis occurs in a 5' to 3' direction.

c. All catalyze the addition of a nucleotide via the germination of a phosphodiester bond.

d. All release pyrophosphate equally a production.

Distinctive features include:

a. The substrates: Dna polymerase, reverse transcriptase, and telomerase employ deoxyribonucleoside triphosphates equally a substrate, whereas RNA polymerase uses ribonucleoside triphosphates.

b. The templates: DNA polymerase and RNA polymerase utilize Dna as a template, whereas telomerase copies an RNA template that is role of the enzyme. Reverse transcriptase uses RNA equally a template in the life cycle of retroviruses and retrotransposons, but in vitro it tin can use either Dna or RNA as a template.

c. Primer requirements: DNA polymerase, reverse transcriptase and telomerase require primers provided past some other activeness or poly peptide (primase, an tRNA or the 3� end of a DNA strand, respectively), whereas RNA polymerase tin begin synthesis of RNA internally to the template without a primer.

In general, the chemistry of the enzyme reaction is similar for all four, but the specific substrates, templates and primers differ.

Answers, Chapter 11. Transcription: Promoters and Terminators

xi.i a) Correct to left

b) 1800

c) 400

5' stop label: The lack of protection of the labeled Kpn-Bgl* 500 nucleotide fragment tells you that the mRNA is synonymous with the lesser strand, and thus the peak strand is the template strand. The tiptop strand is labeled at the 5' stop of the 1500 nucleotide * Bgl-Kpn fragment, and hybridization of this probe with mRNA gives protection of a 1300 nt fragment. This indicates that transcription gain from right to left (on the map every bit given), and the v'' end of the transcript is 1300 nts to the right of the Bgl II site. In the coordinates of the map, this would be 500 (position of Bgl II) + 1300 = 1800.

five'Kpn 500 Bgl Bgl 1500 Kpn

---------------------> *--------------------------------------->

<--------------------* <---------------------------------------

� �

no protection

5' Bgl Kpn

*--------------------------------------->

| | | | | | | | | | | | | | | | | | | | | | | | |

<~~~~~~~~~~~~~~~~~~~~~~~~~�cap

� S1

� gel

*--------------------------->

1300 nt protected fragment

When the five' end label is at the Eco RI site, a similar result is obtained, but one can map the v' stop with greater accuracy. The protected fragment is from the pinnacle strand and is 100 nts (a size that can be measured more accurately than the 1300 nt fragment on the polyacrylamide gels used in this analysis).

Kpn 1700 Eco Eco 300 Kpn

*--------------------------------------------> *----------------->

<--------------------------------------------* ------------------

�

*----------------->

| | | | |

<~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~�

� S1,..etc.

100 nt

*------->

iii' end characterization: When the Deoxyribonucleic acid fragments are labeled at the iii' end, over again the top strand will exist protected by hybridization to mRNA, thus reaffirming the conclusions above that the top strand is the template strand. The 500 nt Kpn-Bgl * fragment generates a 100 nt protected fragment, showing that the 3' end of the mRNA is 100 nts to the left of the Bgl II site, or at 500 - 100 = 400 on the coordinates of the map.

five'Kpn 500 3' Bgl Bgl 1500 Kpn

----------------->* ------------------------------------>

----------------- *------------------------------------

� �

five' -----------------* no protection

| | | |

<~~~~~~~~~~~~~~~~~~~~�

� S1, gel,..etc.

-----*

100 nt

eleven.ii a) Left to right.

b) 400

c) Cannot be determined.

11.3 a) It will increase expression of the almond cistron.

b) It has no effect.

c) It volition increase expression of the almond factor.

d) The -fifty to -i fragment is acting like a promoter. In the commencement set of experiments, it is needed for promotion of transcription and information technology is needed to respond to upstream activating sequences. In the second set of experiments, the heterologous promoter will substitute for it.

xi.4 a) Two complexes are formed betwixt the labeled probe and the kidney cell nuclear extract.

b) Lanes 3-8 tell you that complexes A and B are specific, i.due east. the proteins are recognizing a item Dna sequence, since the self-DNA competes, but the E. coli Dna does non.

c) Lanes ix-14 tell you that the protein binding to grade complex A will as well bind to a Deoxyribonucleic acid containing an Sp1 binding site. Thus the protein that forms complex A with this probe may be Sp1 or a relative of this protein. Neither protein (for circuitous A or complex B) will bind to the Dna probe with the Oct1 binding site, showing that this is non a candidate for the poly peptide forming the sequence-specific complexes with the probe.

11.five five' GAGTC

3' CTCAG

11.six a) False

b) True

c) True

d) True

Answers to questions in Affiliate 12. RNA Processing

12.1 a) NTP labeled at either the b or the g position.

b) NTP labeled at the b position, which will exist in the cap construction.

c) NTP labeled at the a position, since the b and k phosphates are lost upon incorporation of the NMP.

12.2 One of the key signals for cleavage and iii' polyadenylation is the sequence AAUAAA. Subsequently RNA polymerase Ii has transcribed across this sequence, an endonuclease (uncharacterized at this time) cleaves the primary transcript at a position about 25 to xxx nucleotides three' to the AAUAAA. Then the enzyme polyadenylate polymerase adds a string of 20 to 250 A'south to the free three' end, generating the 3' poly(A) tail.

The mutation would prevent cleavage and polyadenylation at the usual site, which could take two dissimilar consequences. If the transcript is not polyadenylated, information technology will exist quite unstable and the steady land levels of mRNA volition be very low, and therefore trivial or no protein product will be made. In some cases, a "substitute AAUAAA" may happen to occur in the transcribed region 3' to the gene, in which case this culling, "ambiguous" polyadenylation site could be used. However, this cryptic site may not be used as efficiently equally the wild type (normal) sequence, again resulting in a reduction in the amount of steady state mRNA.

12.3 a) False

b) Truthful

c) True

d) True

12.4 Both are capable of self-splicing, and both use a phosphoester transfer mechanism (transesterification) that is initiated by a guanine nucleoside or nucleotide. Neither require ATP hydrolysis.

12.five a) Introns almost invariably have the dinucleotides GU at their 5' end and AG at their iii' finish.

b) U1 snRNP

c) A branch structure forms by linkage between the 2' hydroxyl of an A at the branch site in the intron and the 5' phosphate at the 5' G of the intron.

d) Spliceosome assembly require hydrolysis of ATP.

12.6 The mechanism for removal of introns from pre‑mRNAs is very similar to that of the Group II introns, with the germination of a lariat intermediate later on the reaction is initiated. Each of the cleavage and rejoining reactions is a transesterification, in which a new phosphodiester bond is formed for every one that is broken.

A minimum of two transesterification steps are required. The first step is initiated by the set on of a ii' hydroxyl of an A inside the intron on the bail linking the three' cease of the beginning exon with the 5' terminate of the intron. This generates a three' hydroxyl on the nucleotide at the 3' end of the commencement exon, and finer takes the intron out of the serial of transesterifications by forming a lariat structure. This 3' nucleotide of the outset intron tin then link to the beginning nucleotide of the 2nd exon, over again by a transesterification. The issue of this second step is the union of the first and second exons, with the intron liberated as a lariat intermediate.

12.seven a) 2, v

b) 3

c) 1, 4

12.8 (a) The excised intron has a co-operative bespeak where an RNA chain is covalently fastened to the original chain of RNA.

(b) The excised intron has a circle and a linear tail, i.e. it is a lariat.

(c) Cleavage in the region complementary to oligo 1 volition linearize the nonlinear structure and generate a molecule that is just about a full-length intron. E.g. it could open up the circular part of a lariat, but it opens it adjacent to the branch or joint.

(d) The combined results with both oligos 3 and 4 show that co-operative betoken in the excised intron is located between the segments complementary to oligodeoxyribonucleotides iii and 4.

(e) The construction of the excised intron is:

Answers to questions from Chapter 13. Genetic Code

thirteen.ane DNA and RNA polymerases take several properties in common.

i. All are template directed, synthesizing a sequence complementary to the template.

ii. Synthesis occurs in a 5' to three' management.

3. All catalyze the addition of a nucleotide via the formation of a phosphodiester bond.

iv. All use (deoxy)ribonucleoside triphosphates equally a substrate, and release pyrophosphate as a product.

The enzyme polynucleotide phosphorylase can exist used to synthesize RNA in vitro , and this was a fundamental technique in deciphering the genetic code. However, it differs from Deoxyribonucleic acid and RNA polymerases in points 1 and 4. Polynucleotide phosphorylase does not utilise a template, but rather adds ribonucleotides to an RNA in a highly reversible reaction. The substrates (in the management of synthesis) are ribonucleoside diphosphates, which are added with the release of phosphate every bit a product. In the cell, this enzyme probably catalyzes the opposite reaction to degrade RNAs.

13.two a) It starts at AUG (nucleotides v-7) and ends at UGA (nucleotides 23-25). 6 amino acids are encoded, including the initiating methionine.

b) All codons have a U at the second position, hence only hydrophobic amino acids are encoded.

13.3 a) More than ane codon encodes an amino acid.

b) third.

c) The base in the 5' position of the anticodon can frequently pair with several bases in the "wobble" or third position of the codon (e.grand. I with C, U, or A). Therefore, one tRNA can recognize several codons.

thirteen.4 The template strand is the strand that serves equally the template for RNA synthesis; nontemplate strand is identical in sequence with the RNA transcribed from the gene, with U in place of T..

a) (5')CGACGGCGCGAAGUCAGGGGUGUUAAG(3')

b) Arg-Arg-Arg-Glu-Val-Arg-Gly-Val-Lys

c) No; The base sequence of mRNA transcribed from the nontemplate strand would be: (5')CUUAACACCCCTGACUUCGCGCCGUCG. This mRNA when translated would result in a different peptide than in (b). The complementary antiparallel strands in double-helical DNA do not have the same base sequence in the five'to iii' direction. RNA is transcribed from only one specific strand of duplex DNA. The RNA polymerase must therefore recognize and bind to the correct strand.

13.five The two DNA codons for Glu are GAA and GAG, and the iv DNA codons for Val are GTT, GTC, GTA, and GTG. A single-base of operations change in GAA to form GTA or in GAG to form GTG could account for the Glu � Val replacement in sickle-jail cell hemoglobin. Much less probable are two-base changes from GAA to GTG, GTT, or GTC; and from GAG to GTA, GTT, or GTC.

xiii.six AAA

13.vii a) CAG

b) UUG

thirteen.8 a) Three tRNAs are required. Anticodons iii' GCI and 3' GCC tin can accommodate the five' CGN codons, and anticodon 3' UCU volition pair with the 5' AGR codons.

b) Two tRNAs are required. The 5' GUN codon can be matched with anticodons iii' CAI + 3' CAC, or iii' CAG + 3' CAU.

thirteen.9 a) Glycine should attach to a tRNAs with codons 5' GGU, 5' GGC and 5' GGA.

b) Isoleucine should attach to tRNAs with codons 5' AUC and v' AUU.

13.x Some amino acids are encoded past 6 different codons, some four different codons, some iii different codons, some 2 different codons, and some one codon. To minimize the degree of ambiguity in codon assignment for a given peptide sequence, one must select a region of the peptide that contains mostly amino acids specified by a small number of codons.

Focus on the amino acids with the fewest codons: Met and Trp. The best possibility is the span of Deoxyribonucleic acid from the codon for the outset Trp residue to the first two nucleotides of the codon for Ile. The sequence of the probe would be:

(v')UGGUA(U/C)UG(U/C)AUGGA(U/C)UGGAU

The synthesis would be designed to comprise either U or C where indicated, producing a mixture of viii twenty-nucleotide probes that differ but at one or more than of these positions.

thirteen.xi a) 2 nucleotides

b) No

c) AA Codon

1 KK

2 NN

three DD

four KN

v NK

6 DK

seven KD

8 ND

d) DN

eastward) DK � NK

aa6 aa5

f) ND � KD

aa8 aa7

Answers to Chapter 14. Translation

14.1 There are ii tRNAs for methionine: tRNAfMet, the initiating tRNA, and tRNAMet, which can insert Met in interior positions in a polypeptide. tRNAfMet reacts with Met to yield Met-tRNAfMet, promoted by methionine aminoacyl-tRNA synthetase. The amino group of its Met residue is and so formylated by Due north 10-formyltetrahydrofolate to yield fMet-tRNAfMet. Free Met or Met-tRNAMet cannot be formylated. Only fMet-tRNAfMet is recognized past the initiation factor IF-2 and is aligned with the initiating AUG positioned at the ribosomal P site in the initiation complex. AUG codons in the interior of the mRNA are eventually positioned at the ribosomal P site and tin bind and contain only Met-tRNAMet.

xiv.two a) True

b) False

c) True

d) True

14.three a) The C terminal peptide.

b) All peptides accept the aforementioned specific action.

14.iv a) EF-G-GTP

b) IF-2, with GTP

c) RF-one

d) The 16S rRNA in the pocket-size ribosomal subunit.

fourteen.5 a) Small or xxx South

b) AGGA in mRNA is complementary to the 3' end of 16S rRNA occurs before the AUG initiation codon.

c) IF3; IF2-f-Met-tRNA-GTP; IF1; 50S ribosomal subunit

14.half-dozen ane) aa + tRNA + ATP � AMP + PPi + aa - tRNA

aa - tRNA synthetase

2) Binding of aa-tRNA to A site on ribosome;

requires EF-Tu and GTP � Gdp + Pi

3) Translocation of peptidyl-tRNA to P site on ribosome;

requires EF-M and GTP � GDP + Pi

xiv.seven The amino acrid most recently added to a growing polypeptide concatenation is the just one covalently attached to a tRNA and hence is the simply link betwixt the polypeptide and the mRNA that is encoding it. A proofreading action would sever this link, halting synthesis of the polypeptide and releasing it from the mRNA.

xiv.8 An East. coli promoter is required for transcription, because East. coli RNA polymerase does non collaborate with eukaryotic promoters; a ribosome bounden site positioned at an appropriate altitude upstream from the ATG codon is required, considering eukaryotic mRNA does not utilize such a site for translation initiation; an operator site is required for regulation of transcription in E. coli .

14.9 I hypothesis would exist that the codon pair AUY has "always" encoded isoleucine, and in early on development, the codon pair AUR encoded methionine. Subsequent specialization in the utilise of AUG has allowed it exclusively to be used to encode methionine, thereby allowing AUA to be recruited as an additional codon for isoleucine. This hypothesis assumes that both isoleucine and methionine were used in proteins early on in evolution.

An culling hypothesis states that AUN was originally a codon family encoding isoleucine and subseqently, the AUG was recruited to encode methionine. This hypothesis requires either that methionine was non used in proteins early on in evolution, or that it was encoded by another codon as well AUG. Neither of these latter possibilities seems very probable, and then the alternative hypothesis is harder to rationalize. Nonetheless, it cannot exist ruled out, since nosotros do non have direct access to find the weather condition of early development.

14.10 (a) one and 5 are right.

(b) 2, four, seven are right.

(c) 3 and 6 are correct.

browningrabliver.blogspot.com

Source: http://www.bx.psu.edu/~ross/workmg/Answers_Part_Three.htm

Share this post