
Concept explainers
(a)
To explain: The process by which the result given in question is consistent with posttranscriptional modification of the AMPA receptor
Introduction:
The molecular method through which the cells make discrete changes to specific
(a)

Explanation of Solution
The complementary cDNA is formed from the mRNA by the process known as reverse transcription of DNA. Here, in this the
(b)
To explain: The reason for the different products that are formed from the edited and unedited
Introduction:
RNA editing means that editing events means insertion, deletion, and base substitution of nucleotides within the edited RNA molecule. Post transcriptional regulation stands for the regulation of gene expression at. It is the process in which primary transcript of RNA is converted into a mature RNA.
(b)

Explanation of Solution
The unedited mRNA sequence remains same that of DNA just thymine is replaced by uracil.
The unedited mRNA has the sequences (site of editing is indicated by
The annealing of primer 1 is shown as follows:
The cDNA is formed from right to left and is as follows:
The resulted cDNA is as follows:
The annealing of primer 2 is shown as follows:
The nucleotides are added to
The nineteen nucleotides are formed for unedited transcript. The
The edited bases are elongated by DNA polymerase and will terminate at consecutive T in cDNA. The newly formed DNA is marked and is as follows:
The DNA polymerase is elongated and the edited base stops at the next thymine residue and gives the product of 22 nucleotides.
(c)
To determine: The key weakness in the argument that the editing machinery consists of protein.
Introduction:
RNA editing means that editing events means insertion, deletion, and base substitution of nucleotides within the edited RNA molecule. Post transcriptional regulation stands for the regulation of gene expression at. It is the process in which primary transcript of RNA is converted into a mature RNA.
(c)

Explanation of Solution
To disrupt the protein function proteases and heat treatment methods are used, this will inhibit the editing event. Nucleases will act only on
This clash revels that the disrupting treatment of protein will not halt the process of editing. In the absence of enzyme, the process of editing and degradation of
(d)
To determine: The importance of using [α-32P] ATP rather than [β-32P] ATP or [γ-32P] ATP in the experiment.
Introduction:
RNA editing means that editing events means insertion, deletion, and base substitution of nucleotides within the edited RNA molecule. Post transcriptional regulation stands for the regulation of gene expression at. It is the process in which primary transcript of RNA is converted into a mature RNA.
(d)

Explanation of Solution
The alpha-phosphate of the nucleoside triphosphate is inserted into the polynucleotide. In this researchers have labeled added phosphates, then none product will have the radio labeled phosphate.
(e)
To determine: The importance of using [α-32P] ATP rather than [β-32P] GTP or [γ-32P] UTP in the experiment.
Introduction:
RNA editing means that editing events means insertion, deletion, and base substitution of nucleotides within the edited RNA molecule. Post transcriptional regulation stands for the regulation of gene expression at. It is the process in which primary transcript of RNA is converted into a mature RNA.
(e)

Explanation of Solution
In the given reaction it is clear that only one adenine residue is edited, so the fate of other adenine residues in the sequence will change and this will change the sequence of the whole strand. So, the adenine residue in sequence becomes important for further studies. These changes will help in understanding the importance of using [α-32P] ATP than [β-32P] GTP or [γ-32P].
(f)
To determine: The process by which the entire
Introduction:
RNA editing means that editing events means insertion, deletion, and base substitution of nucleotides within the edited RNA molecule. Post transcriptional regulation stands for the regulation of gene expression at. It is the process in which primary transcript of RNA is converted into a mature RNA.
(f)

Explanation of Solution
The ATP is only labeled, if the whole nucleotide were removed all type of radioactivity have been detached from the mRNA. Since, the unmodified AMP containing P32 is present in the chromatographic plate.
(g1)
To determine: The effect of result that exclude removal of the A base followed by replacement with I base as a mechanism of editing.
Introduction:
RNA editing means that editing events means insertion, deletion, and base substitution of nucleotides within the edited RNA molecule. Post transcriptional regulation stands for the regulation of gene expression at. It is the process in which primary transcript of RNA is converted into a mature RNA.
(g1)

Explanation of Solution
In the presence of AMP, if any base is replaced. This indicates that A has changed to I. This alteration occurs without the removal of
If the bases were removed or replaced then only AMP will appear but the presence of IMP indicates that base A is replaced with I base. This replacement takes place without the removal of hydrogen. So if A is not replaced by I then editing will not occur and AMP will not change in IMP.
(g2)
To determine: The mechanism behind the editing process in the replacement of
Introduction:
RNA editing means that editing events means insertion, deletion, and base substitution of nucleotides within the edited RNA molecule. Post transcriptional regulation stands for the regulation of gene expression at. It is the process in which primary transcript of RNA is converted into a mature RNA.
(g2)

Explanation of Solution
In the presence of AMP, if any base is replaced. This indicates that A has changed to I. This alteration occurs without the removal of
(h)
To determine: The change that occur in Gln to Arg in the protein sequences in the two forms of AMPA receptor.
Introduction:
RNA editing means that editing events means insertion, deletion, and base substitution of nucleotides within the edited RNA molecule. Post transcriptional regulation stands for the regulation of gene expression at. It is the process in which primary transcript of RNA is converted into a mature RNA.
(h)

Explanation of Solution
The codon that will change under this process is CAG. This CAG will convert into CIG. This codon is known as CGG.
Want to see more full solutions like this?
Chapter 26 Solutions
EBK LEHNINGER PRINCIPLES OF BIOCHEMISTR
- write the ionization equilibrium for cysteine and calculate the piarrow_forwardplease answerarrow_forwardf. The genetic code is given below, along with a short strand of template DNA. Write the protein segment that would form from this DNA. 5'-A-T-G-G-C-T-A-G-G-T-A-A-C-C-T-G-C-A-T-T-A-G-3' Table 4.5 The genetic code First Position Second Position (5' end) U C A G Third Position (3' end) Phe Ser Tyr Cys U Phe Ser Tyr Cys Leu Ser Stop Stop Leu Ser Stop Trp UCAG Leu Pro His Arg His Arg C Leu Pro Gln Arg Pro Leu Gin Arg Pro Leu Ser Asn Thr lle Ser Asn Thr lle Arg A Thr Lys UCAG UCAC G lle Arg Thr Lys Met Gly Asp Ala Val Gly Asp Ala Val Gly G Glu Ala UCAC Val Gly Glu Ala Val Note: This table identifies the amino acid encoded by each triplet. For example, the codon 5'-AUG-3' on mRNA specifies methionine, whereas CAU specifies histidine. UAA, UAG, and UGA are termination signals. AUG is part of the initiation signal, in addition to coding for internal methionine residues. Table 4.5 Biochemistry, Seventh Edition 2012 W. H. Freeman and Company B eviation: does it play abbreviation:arrow_forward
- Answer all of the questions please draw structures for major productarrow_forwardfor glycolysis and the citric acid cycle below, show where ATP, NADH and FADH are used or formed. Show on the diagram the points where at least three other metabolic pathways intersect with these two.arrow_forwardanswer the questions please all of them should be answeredarrow_forward
- Burk plot is shown below. Calculate Km and max for this enzyme. show workarrow_forwardInsert Format Tools Extensions Help Normal text ▾ Arial C 2 10 3 + BIUA Student Guide (continued) Record data and conclusions about the mystery food sample either below or in a lab notebook. Step 2: Protein Test (Biuret Solution) Gelatin Water [Mystery Food (Positive Control) (Negative Control) Sample pink purple no change no change They mystery food sample does not contain protein because the color of the test tube wasn't pink or purple Color Conclusion They mystery food sample does not contain protein because the color of the test tube wasn't pink or purple Step 3: Lipid Test (Sudan Red Solution) Vegetable Oil Water (Positive Control) (Negative Control) Mystery Food Sample floating red no change floating red the mystery food dosnt contain lipids because the test tube has floating red 75 % 87 8 9 7 ChromeOS C Device will pow 26.battery lea powerarrow_forwardThe rate data from an enzyme catalyzed reaction with and without an inhibitor present is found in the image. Question: what is the KM and Vm and the nature of inhibitionarrow_forward
- 1. Estimate the concentration of an enzyme within a living cell. Assume that: (a): fresh tissue is 80% water and all of it is intracellular (b): the total soluble protein represents 15% of the weight (c): all the soluble proteins are enzymes (d): the average molecular weight of the proteins is 150,000 (E): about 100 different enzymes are present please help I am lostarrow_forwardPlease helparrow_forwardThe following data were recorded for the enzyme catalyzed conversion of S -> P. Question: Estimate the Vmax and Km. What would be the rate at 2.5 and 5.0 x 10-5 M [S] ?arrow_forward
- BiochemistryBiochemistryISBN:9781319114671Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.Publisher:W. H. FreemanLehninger Principles of BiochemistryBiochemistryISBN:9781464126116Author:David L. Nelson, Michael M. CoxPublisher:W. H. FreemanFundamentals of Biochemistry: Life at the Molecul...BiochemistryISBN:9781118918401Author:Donald Voet, Judith G. Voet, Charlotte W. PrattPublisher:WILEY
- BiochemistryBiochemistryISBN:9781305961135Author:Mary K. Campbell, Shawn O. Farrell, Owen M. McDougalPublisher:Cengage LearningBiochemistryBiochemistryISBN:9781305577206Author:Reginald H. Garrett, Charles M. GrishamPublisher:Cengage LearningFundamentals of General, Organic, and Biological ...BiochemistryISBN:9780134015187Author:John E. McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. PetersonPublisher:PEARSON





