Concept explainers
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(a)
Interpretation:
The initiation sequence, a control segment of DNA has to be accounted.
Concept Introduction:
Composition of
RNA synthesis: The process of RNA synthesis is Transcription. A small section of DNA unwinds, only one of the two strands act as template and the other strand as informational strand. The complementary bases are attached one by one by the action of RNA polymerase at template strand on moving down. The newly generated RNA is the exact copy of the informational strand, with the exception that a U replaces each T in the template DNA. The RNA synthesised carries genetic information and directs protein synthesis.
Replication of DNA: The process by which copies of DNA are made when a cell divides.
(b)
Interpretation:
The informational strand of DNA has to be predicted.
Concept Introduction:
Composition of nucleic acid: Nucleic acid is a polymer of nucleotides. Each nucleotide has three parts: a sugar, a nitrogenous base, and a phosphate group. Two nucleotides are joined by phosphate diester linkage where a free phosphate on 5’ carbon of one nucleotide and a free –OH group on 3’ carbon of another nucleotide is linked.
Replication of DNA: The process by which copies of DNA are made when a cell divides.
RNA synthesis: The process of RNA synthesis is Transcription. A small section of DNA unwinds, only one of the two strands act as template and the other strand as informational strand. The complementary bases are attached one by one by the action of RNA polymerase at template strand on moving down. The newly generated RNA is the exact copy of the informational strand, with the exception that a U replaces each T in the template DNA. The RNA synthesised carries genetic information and directs protein synthesis.
(c)
Interpretation:
The template strand of DNA has to be accounted.
Concept Introduction:
Composition of nucleic acid: Nucleic acid is a polymer of nucleotides. Each nucleotide has three parts: a sugar, a nitrogenous base, and a phosphate group. Two nucleotides are joined by phosphate diester linkage where a free phosphate on 5’ carbon of one nucleotide and a free –OH group on 3’ carbon of another nucleotide is linked.
RNA synthesis: The process of RNA synthesis is Transcription. A small section of DNA unwinds, only one of the two strands act as template and the other strand as informational strand. The complementary bases are attached one by one by the action of RNA polymerase at template strand on moving down. The newly generated RNA is the exact copy of the informational strand, with the exception that a U replaces each T in the template DNA. The RNA synthesised carries genetic information and directs protein synthesis.
Replication of DNA: The process by which copies of DNA are made when a cell divides.
(d)
Interpretation:
The end of the sequence (termination sequence) has to be accounted.
Concept Introduction:
Composition of nucleic acid: Nucleic acid is a polymer of nucleotides. Each nucleotide has three parts: a sugar, a nitrogenous base, and a phosphate group. Two nucleotides are joined by phosphate diester linkage where a free phosphate on 5’ carbon of one nucleotide and a free –OH group on 3’ carbon of another nucleotide is linked.
RNA synthesis: The process of RNA synthesis is Transcription. A small section of DNA unwinds, only one of the two strands act as template and the other strand as informational strand. The complementary bases are attached one by one by the action of RNA polymerase at template strand on moving down. The newly generated RNA is the exact copy of the informational strand, with the exception that a U replaces each T in the template DNA. The RNA synthesised carries genetic information and directs protein synthesis.
Replication of DNA: The process by which copies of DNA are made when a cell divides.
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Chapter 26 Solutions
EP FUND.OF GENERAL,ORG...-MOD.MASTERING
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- Draw a diagram to demonstrate 3' to 5' exonuclease activity?arrow_forward1. Write the piece of mRNA that would code for the amino acid sequence NH3 - met - tyr - cys- his- CO2. Label the 5' and 3' ends. (diagram attached) 2.arrow_forwardYou were tasked with purifying an untagged transcription factor (molecular weight 65,000Da, isoelectric point unknown) from a contaminant Protein A (molecular weight 50,000 Da, isoelectric point 5.0). You were also instructed to use the protein for crystallization studies after purification.i) In the initial purification step, which type of chromatography should you attempt? Explain your choice and specify the requirements of the buffer solution you should use. ii) Analyzing the proteins recovered from step i) using SDS-PAGE, you still observed a faint band of Protein A in addition to the transcription factor band. Given the limited time available for further purification, you must choose ONE appropriate chromatography method to maximize the chances of separating the two proteins as well as your crystallization studies. Provide a detailed explanation of your selection and the techniques/strategies you would employ to achieve this.arrow_forward
- You were given a mixture of two proteins with different isoelectric points and molecular weights:• Protein X: pI 4.2, MW 42,000• Protein Y: pI 9.8, MW 90,000Using a Tris-glycine discontinuous native gel (pH8.3) electrophoresis system with a running gel of 12%, only a single band was observed upon protein staining after electrophoresis. Explain the observed result and discuss possible factors affecting protein migration in this system.arrow_forwardThe standard cost of Product B manufactured by Oriole Company includes 3.5 units of direct materials at $5.40 per unit. During June, 27,300 units of direct materials are purchased at a cost of $5.15 per unit, and 27,300 units of direct materials are used to produce 7,600 units of Product B. (a) Compute the total materials variance and the price and quantity variances. Total materials variance Materials price variance Materials quantity variance $ (b) Compute the total materials variance and the price and quantity variances, assuming the purchase price is $6.35 and the quantity purchased and used is 26,300 units. Total materials variance Materials price variance Materials quantity variance $arrow_forwardThe pyruvate dehydrogenase complex catalyzes the oxidative decarboxylation of pyruvate to form acetyl CoA. E₁, E2, and E3 are abbreviations for the enzymes of the complex. Classify the enzyme names, prosthetic groups, and reactions as E1, E2, or E3. E₁ E2 Answer Bank E3 transfer of electrons to FAD and then to NAD+ transfer of acetyl group to coenzyme A formation of hydroxyethyl-TPP hydroxyethyl group transferred to lipoamide thiamine pyrophosphate (TPP) FAD lipoamide dihydrolipoyl transacetylase pyruvate dehydrogenase dihydrolipoyl dehydrogenasearrow_forward
- Patients with pyruvate dehydrogenase deficiency show high levels of lactic acid in the blood. However, in some cases, treatment with dichloroacetate (DCA), which inhibits the kinase associated with the pyruvate dehydrogenase complex, lowers lactic acid levels. How does DCA act to stimulate pyruvate dehydrogenase activity? DCA activates pyruvate dehydrogenase kinase. DCA increases phosphorylation levels of pyruvate dehydrogenase. DCA inhibits pyruvate dehydrogenase kinase. ODCA activates pyruvate dehydrogenase phosphatase. What does this suggest about pyruvate dehydrogenase activity in patients who respond to DCA? The pyruvate dehydrogenase complex is active only when phosphorylated by the kinase. The pyruvate dehydrogenase complex is active only in the presence of the kinase. The pyruvate dehydrogenase complex is completely inactive. The pyruvate dehydrogenase complex displays some residual activity.arrow_forwardThe reduced coenzymes generated by the citric acid cycle donate electrons in a series of reactions called the electron-transport chain. The energy from the electron-transport chain is used for oxidative phosphorylation. Which compounds donate electrons to the electron- transport chain? H₂O NADH பப NAD+ ATP ADP FADH₂ FAD Which compounds are the final products of the electron-transport chain and oxidative phosphorylation? H₂O NADH NAD+ ΠΑΤΡ Π ADP FADH₂ FAD Which compound is the final electron acceptor in the electron-transport chain? Оно NADH NAD+ ATP ADP FADH₂ FADarrow_forwardHexokinase in red blood cells has a Michaelis constant (KM) of approximately 50 μM. Because life is hard enough as it is, let's assume that hexokinase displays Michaelis-Menten kinetics. What concentration of blood glucose yields an initial velocity (V) equal to 90% of the maximal velocity (Vmax)? [glucose] = What does the calculated substrate concentration at 90% Vmax tell you if normal blood glucose levels range between approximately 3.6 and 6.1 mM? Hexokinase operates near Vmax only when glucose levels are low. Hexokinase normally operates far below Vmax. Hexokinase operates near Vmax only when glucose levels are high. Hexokinase normally operates near Vmax mMarrow_forward
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