14 A. Identify the major steps that could account for the 45 minute lag in appea
ID: 51304 • Letter: 1
Question
14
A. Identify the major steps that could account for the 45 minute lag in appearance of calbindin.
B.Eucaryotic cells (cells without a nucleus)attach amino acids to new proteins at the rate of
about 2 per second. Is the synthesis of the protein the major par
t of the lag? (Hint: the average molecular weight of an amino acid is about 100 daltons. You can estimate how
many amino acids are in the protein from this information-you could look it up because itssequence is known, but we are just doing a “back of the envelope” calculation here)
.14. We have a double-stranded DNA segment of 1000 base pairs. Its nucleotide composition is
randomly distributed among A, T, C and G. Assume that each hydrogen bond in the double strand
has an energy of 4 kcal mol-1(1 joule = 0.239 cal).
A.How many hydrogen bonds are there in the segment? (Hint: consult Fig. 2.2.3 for the
numbers of hydrogen bonds for each base pair)
B.What is the total energy necessary to pry apart the two strands, assuming that hydrogen
bonding is the only force keeping them together? (It isn’t).
C.Assume that hydrogen bonds break when they are stretched 0.2 nm. How much force is
necessary to break one?
Explanation / Answer
Calbindin refers to several calcium-binding proteins. They were originally described as vitamin D-dependent calcium-binding proteins in the intestine and kidney in the chick and mammals.
calcium transport (or absorption) may be regarded as a three-step process, consisting of
1)entry into the epithelial cell from the lumen;
2) transit through the cytosol, from the apical to the basolateral pole and
3) extrusion from the cell, across the baso lateral membranes, to the vascular supply in the lamina propria.
Transcription of the ~a1bindin-D~~~ gene is directly correlated to nuclear uptake and binding of 1,25-(OH)& to its receptor. These events are followed by the accumulation of the mature message in the cytosol and its subsequent translation to calbindin-DBK. The long lag times before these latter two events occur may reflect the involvement of posttranscriptional mechanisms in the regulation of calbindinDBK gene expression. Presumably, transcription of the calbindin-DfSK gene results from a direct interaction of the occupied 1,25-(OH)2D3 receptor with specific regulatory regions of the gene
The difficulty in relating alkaline phosphatase activity to 1,25(OH)2Dstimulated calcium entry or absorption is the relatively long time lag associated with the response (1). How ever, Bachelet et al. observed a significant increase in enzyme activity only 30 min after the onset of 1,25(OH)2D infusion into the intestinal blood supply. More recently, Nasr et al. observed an early and biphasic response to a single intraperitoneal injection of 1,25(OH)2D to vitamin D-deficient rats. Brush bor der alkaline phosphatase activity was significantly en hanced at 10 min postdose, peaked at 45 min, declined at l h and gradually increased again over the ensuing 8-h period. These results provide a temporal coinidence between enhanced brush border alkaline phos phatase activity and calcium entry.
B. yes,
synthesis of the protein the major part of the lag
Explanation:The initial codon recognition, however, triggers the elongation factor to hydrolyze its bound GTP (to GDP and inorganic phosphate), whereupon the factor dissociates from the ribosome without its tRNA, allowing protein synthesis to proceed. The elongation factor introduces two short delays between codon-anticodon base pairing and polypeptide chain elongation; these delays selectively permit incorrectly bound tRNAs to exit from the ribosome before the irreversible step of chain elongation occurs. The first delay is the time required for GTP hydrolysis. The rate of GTP hydrolysis by EF-Tu is faster for a correct codon-anticodon pair than for an incorrect pair; hence an incorrectly bound tRNA molecule has a longer window of opportunity to dissociate from the ribosome. In other words, GTP hydrolysis selectively captures the correctly bound tRNAs. A second lag occurs between EF-Tu dissociation and the full accommodation of the tRNA in the A site of the ribosome. Although this lag is believed to be the same for correctly and incorrectly bound tRNAs, an incorrect tRNA molecule forms a smaller number of codon-anticodon hydrogen bonds than does a correctly matched pair and is therefore more likely to dissociate during this period. These two delays introduced by the elongation factor cause most incorrectly bound tRNA molecules (as well as a significant number of correctly bound molecules) to leave the ribosome without being used for protein synthesis, and this two-step mechanism is largely responsible for the 99.99% accuracy of the ribosome in translating proteins.
C.however the question is incomplete
Still possible solution is:
leyt say, from the literature, Homosapiens has 41% GC content,and 59% AT composition
therefore, in 1000 base pairs, 410 will be CG and 590 will be AT.
there are two hydrogen bonds between Aand T so, total AT hydrogen bonds= 590X2= 1180
and hydrogen bonds between GC= 410X3=1230
total no. of hydrogen bonds= 1180+1230=2410
B. one hydrogen bond=4kcal/mol
therefore, total energy = 2410X4= 9640Kcal/mol.
energy greater than 9640KCal/mol will be required to pry apart the two strands.
C. 1 joule = 0.239 cal
energy required to break the bonds= 9640Kcal/mol
in joules,9640Kcal/mol=6.9477×1021X9640= 66975.828X10-21joules/molecule
force= work done/distance
=6.6975X10-17/0.2= 3.3487X10-17J/nm=3.3487X10-7 Nm is the force required to break all hydrogen bonds.
so for one bond breakage force requitred,= 3.3487X10-7/2410= 1.389X1010-11 Nm.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.