The essential attributes of all living things include all EXCEPT ________. Program Improvisation Compartmentalization Emotion Which recognition sequence is NOT a palindrome? (DNA sequences are shown in the standard 5'→ 3' direction) AAATTT GGACCC TCTAGA AAGCTT The standard reduction potential is a measure of the ability of an oxidant to ________ an electron. Lose Multiply Divide Gain Proteins in biological membranes ________. May be porous May not be attached to the membrane surface May not span the membrane May be integrally-bound only At the midpoint of a titration curve ________. The concentration of a conjugate base is 1/2 that of the concentration of a conjugate acid The ability of the solution to buffer is at its least effective The concentration of a conjugate base is twice that of the concentration of a conjugate acid The pH equals the pKa ________ has permitted rapid advances in our understanding of structural macromolecules from living cells. Biokinetics Bioinformatics Biothermodynamics Biogenesis Amphipathic molecules are ________; not able to interact via van der Waals forces. Nonpolar only Polar only Neither polar nor nonpolar Both polar and nonpolar A reaction with a ________ free energy of hydrolysis can be coupled to the phosphorylation of ADP and Pi to ATP. Large negative Neutral Large positive Small negative If the standard reduction potential of a certain half-reaction is -0.30 V, which statement below is TRUE for this half-reaction to proceed as a reduction? This half-reaction will proceed if it is coupled to another half-reaction that has a standard reduction potential greater than -0.30 V. This half-reaction will proceed if it is coupled to another half-reaction that has a standard reduction potential greater than +0.30 V. This half-reaction will proceed if it is coupled to another half-reaction that has a standard reduction potential less than -0.30 V. It doesn't matter what half-reaction this is coupled with; it will proceed spontaneously. Which cellular component carries the secretion and transport of biosynthesized proteins? Mitochondria Nucleus Golgi complex Lysosomes The recognition site of some restriction enzymes are shown. Which will produce sticky ends? SmaI: CCCGGG XhoI: CTCGAG EcoRI: GAATTC Both EcoRI and XhoI
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Which cellular molecule will have an antagonistic action to the effects of adenylyl cyclase? - glycogen phosphorylase - Galpha-GTP - phosphodiesterase - cAMP - None of the above A defective hepatocyte was isolated from a liver cancer patient. This hepatocyte contains approximately 75+ active protein kinase A (PKA) molecules, even in the absence of epinephrine. What is a plausible explanation? - The cell's adenylate cyclase gene has a mutation that makes the cyclase inactive all of the time. - The cell's phosphorylase kinase gene has a mutation that makes the kinase active all of the time. - The cell's adenylate cyclase gene has a mutation that makes the cyclase active all of the time. - The cell's glycogen phosphorylase gene has a mutation that makes the phosphorylase active all of the time. - None of the above. The carbon that enters the carbon assimilation reactions of photosynthesis is __________ to form triose phosphates - oxidized by ATP - reduced by ATP - reduced by NADPH - oxidized by NADPH - reduced and phosphorylated by NADPH Microtubules are involved in all of the following EXCEPT - the formation of the spindle apparatus during cell division - the extension of the lamellopodium of a white blood cell during cell locomotion - the formation of flagella used by sperm cells to swim - transport of vesicles between organelles of the endomembrane system - Microtubules are involved in all of the above. Which of the following is NOT true concerning the chloroplasts and mitochondria and therefore is not evidence to support the endosymbiotic theory? - Both are surrounded by a double membrane. - Both divide by binary fission. - Both contain their own circular genome. - Both can survive independently from a eukaryotic cell. - All of the above are TRUE, and support the endosymbiotic theory. After PIP2 is cleaved, where will you find DAG and IP3? - Both remain in the plasma membrane - DAG remains in the plasma membrane, IP3 diffuses through the cytoplasm - IP3 remains in the plasma membrane, DAG diffuses through the cytoplasm - Both diffuse through the cytoplasm What could you add to your in vitro system (above) to stop the MTs from treadmilling? - tubulin-GTP dimers - tubulin-GDP dimers - GDP - Could add either A or B Which of the following is NOT true concerning the chloroplasts and mitochondria and therefore is not evidence to support the endosymbiotic theory? - Both are surrounded by a double membrane. - Both divide by binary fission. - Both contain their own circular genome. - Both can survive independently from a eukaryotic cell. - All of the above are TRUE, and support the endosymbiotic theory.
Md.Daniyal A.
For the following questions, consider these known facts about the role of cI protein in regulating the lysis/lysogeny decision in lambda phage: 1. Binding of cI protein to O_R1 inhibits transcription of the Cro gene. Thus, cI acts as a repressor of Cro gene transcription. 2. Interactions between cI protein bound to O_R2 and RNA polymerase bound to O_R3 (which also serves as the promoter for cI gene transcription) activate transcription of the cI gene. Thus, cI acts as an activator of cI gene transcription. 3. Transcription of the cI gene is inhibited by cI protein bound to O_R3. However, since the affinity of cI for O_R3 is weak, O_R3 is occupied by cI protein only when there is a high concentration of cI protein in the cell. E. If there is a burst of cI gene expression, which function of cI protein will be observed first? Why? F. What advantage might transcriptional inhibition confer when cI protein binds to O_R3? Why? cI protein exists as a dimer in solution and when bound to DNA (see Figure 3). cI dimers bind cooperatively to adjacent operator sites, one repressor dimer binding to each site. The term "cooperativity" refers in this case to a positively cooperative interaction in which the binding of a cI dimer to the high affinity OR1 site increases the affinity of a second dimer for the weaker OR2 site. Thus although the OR1 and OR2 sites differ in their affinities for cI by ~10-fold, two cI dimers bind simultaneously when OR1 and OR2 sites are adjacent (Figure 3). G. How might lambda phage benefit from the positively cooperative binding of cI to OR1 and OR2? H. cI proteins have two domains – the N-terminal domain, which contacts DNA, and the C-terminal domain, which mediates dimer formation and the cooperative interaction between dimers (Figure 3). What would happen to cI and Cro gene expression if the only available cI protein was truncated before the C-terminal domain, such that only the N-terminal domain remained? Why? Under normal circumstances, OR1 and OR2 are adjacent on the DNA, thus the C-terminal domains of a cI dimer bound to OR1 can interact with the C-terminal domains of a dimer bound to OR2 as shown schematically in Figure 3. Hochschild and Ptashne decided to ask what would happen to cooperative binding if the distance between OR1 and OR2 was increased.
Madhur L.
Figure $3-24$ shows the amino acid sequence of bovine insulin. This structure was determined by Frederick Sanger and his coworkers. Most of this work is described in a series of articles published in the Biochemical Journal from 1945 to 1955. When Sanger and colleagues began their work in $1945,$ it was known that insulin was a small protein consisting of two or four polypeptide chains linked by disulfide bonds. Sanger's team had developed a few simple methods for studying protein sequences. Treatment with FDNB. FDNB (1-fluoro-2,4-dinitrobenzene) reacted with free amino (but not amide or guanidinium) groups in proteins to produce dinitrophenyl (DNP) derivatives of amino acids: Acid Hydrolysis. Boiling a protein with $10 \%$ HCl for several hours hydrolyzed all of its peptide and amide bonds. Short treatments produced short polypeptides; the longer the treatment, the more complete the breakdown of the protein into its amino acids. Oxidation of Cysteines. Treatment of a protein with performic acid cleaved all the disulfide bonds and converted all Cys residues to cysteic acid residues (see Fig. $3-28$ ). Paper Chromatography. This more primitive version of thin-layer chromatography (see Fig. $10-25$ ) separated compounds based on their chemical properties, allowing identification of single amino acids and, in some cases, dipeptides. Thin-layer chromatography also separates larger peptides. As reported in his first paper (1945), Sanger reacted insulin with FDNB and hydrolyzed the resulting protein. He found many free amino acids, but only three DNP-amino acids: $a-$ DNP-glycine (DNP group attached to the $\alpha$ -amino group), $a$ -DNP-phenylalanine, and $\varepsilon$ DNP-lysine (DNP attached to the $\alpha$ -amino group). Sanger interpreted these results as showing that insulin had two protein chains: one with Gly at its amino terminus and one with Phe at its amino terminus. One of the two chains also contained a Lys residue, not at the amino terminus. He named the chain beginning with a Gly residue "A" and the chain beginning with Phe "B." (a) Explain how Sanger's results support his conclusions. (b) Are the results consistent with the known structure of bovine insulin (see Fig. 3-24) ? In a later paper ( $1949)$, Sanger described how he used these techniques to determine the first few amino acids (amino-terminal end) of each insulin chain. To analyze the B chain, for example, he carried out the following steps: 1. Oxidized insulin to separate the A and B chains. 2. Prepared a sample of pure B chain with paper chromatography. 3. Reacted the B chain with FDNB. 4. Gently acid-hydrolyzed the protein so that some small peptides would be produced. 5. Separated the DNP-peptides from the peptides that did not contain DNP groups. 6. Isolated four of the DNP-peptides, which were named B1 through B4. 7. Strongly hydrolyzed each DNP-peptide to give free amino acids. 8. Identified the amino acids in each peptide with paper chromatography. The results were as follows: B1: $ \alpha$ -DNP-phenylalanine only B2: $ \alpha$ -DNP-phenylalanine; valine B3: aspartic acid; $\alpha$ -DNP-phenylalanine; valine B4: aspartic acid; glutamic acid; $a$ -DNP-phenylalanine; valine (c) Based on these data, what are the first four (amino-terminal) amino acids of the B chain? Explain your reasoning. (d) Does this result match the known sequence of bovine insulin (Fig. $3-24$ )? Explain any discrepancies. Sanger and colleagues used these and related methods to determine the entire sequence of the A and B chains. Their sequence for the A chain was as follows: Because acid hydrolysis had converted all Asn to Asp and all Gln to Glu, these residues had to be designated Asx and Glx, respectively (exact identity in the peptide unknown). Sanger solved this problem by using protease enzymes that cleave peptide bonds, but not the amide bonds in Asn and Gln residues, to prepare short peptides. He then determined the number of amide groups present in each peptide by measuring the $\mathbf{N H}_{4}^{+}$ released when the peptide was acid-hydrolyzed. Some of the results for the A chain are shown below. The peptides may not have been completely pure, so the numbers were approximate - but good enough for Sanger's purposes. (e) Based on these data, determine the amino acid sequence of the A chain. Explain how you reached your answer. Compare it with Figure $3-24$. (TABLE CAN'T COPY)(EQUATION CAN'T COPY)
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