Biochemistry
9th Edition
ISBN: 9781319114671
Author: Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher: W. H. Freeman
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Chapter 26, Problem 38P
Interpretation Introduction
Interpretation:
The reason for the modification of proteins by the covalent linking of a farnesyl or a geranylgeranyl unit to the carboxyl-terminal cysteine residue should be determined.
Concept introduction:
Phospholipids are the basic constituent of all cell membranes. They usually form lipid bilayers. The structure of a phospholipid molecule has a tail made up of two hydrophobic fatty acid and a hydrophilic head comprising of a phosphate group.
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Students have asked these similar questions
A cytosolic protein has an important alpha amino group. The pKa of this group is approximately 8 when exposed to water outside of a protein.
1. What would happen to the pKa if this group was instead buried in the hydrophobic interior of the protein? Explain.
2. Let’s say in the hydrophobic interior of the protein, the group forms an ionic bond with a carboxylate group of the side chain of a charged Asparagine residue. How would the pKa of this alpha amino group compare with the pKa of the alpha amino group in the hydrophobic interior of the protein without a nearby Asparagine residue to form this ionic bond? Explain.
Glycosylation is a major type of protein post-translational modification. Identify the amino acid that is joined to each
monosaccharide by a glycosidic bond.
glycoprotein A
glycoprotein B
glycoprotein C
HA
HO
Н
CH₂OH
HO
OH Н
H
CH₂OH
Н
ОН
HO
HN-C-CH3
о
c=0
NHI
-NH-C - CH2C-H
ОН
Н
нн
ОН НН
CH2OH
Он
OH HO
Н
NH
1
С=0
-CH2-C-н
NH
C=0
LO-CH2-C-H
NH
A
В
с
The core of pectin molecules is a polymer of a (1®4)-linked D-galacturonate. Draw one of its residues.
Chapter 26 Solutions
Biochemistry
Ch. 26 - Prob. 1PCh. 26 - Prob. 2PCh. 26 - Prob. 3PCh. 26 - Prob. 4PCh. 26 - Prob. 5PCh. 26 - Prob. 6PCh. 26 - Prob. 7PCh. 26 - Prob. 8PCh. 26 - Prob. 9PCh. 26 - Prob. 10P
Ch. 26 - Prob. 11PCh. 26 - Prob. 12PCh. 26 - Prob. 13PCh. 26 - Prob. 14PCh. 26 - Prob. 15PCh. 26 - Prob. 16PCh. 26 - Prob. 17PCh. 26 - Prob. 18PCh. 26 - Prob. 19PCh. 26 - Prob. 20PCh. 26 - Prob. 21PCh. 26 - Prob. 22PCh. 26 - Prob. 23PCh. 26 - Prob. 24PCh. 26 - Prob. 25PCh. 26 - Prob. 26PCh. 26 - Prob. 27PCh. 26 - Prob. 28PCh. 26 - Prob. 29PCh. 26 - Prob. 30PCh. 26 - Prob. 31PCh. 26 - Prob. 32PCh. 26 - Prob. 33PCh. 26 - Prob. 34PCh. 26 - Prob. 35PCh. 26 - Prob. 36PCh. 26 - Prob. 37PCh. 26 - Prob. 38PCh. 26 - Prob. 39PCh. 26 - Prob. 40PCh. 26 - Prob. 41PCh. 26 - Prob. 42PCh. 26 - Prob. 43PCh. 26 - Prob. 44PCh. 26 - Prob. 45PCh. 26 - Prob. 46PCh. 26 - Prob. 47P
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- Globular proteins with multiple disulfide bonds must be heated longer and at higher temperature to denature them. Bovinepancreatic trypsin inhibitor (BPTI), having 58 amino acids in a single chain and 3 disulfide linkages, loses its catalytic activity whenheated at nearly 90°C for 5-10 minutes. Explain the molecular basis of this observed thermal property of BPTI relative to the nativestructure and function of the protein.arrow_forwardeading list Cells have oligosaccharides displayed on their cell surface that are important for cell-cell recognition. Your friend has discovered a transmembrane glycoprotein, GP1, on a pathogenic fungal cell that is recognized by human immune cells. He decides to purify large amounts of GP1 by expressing it in bacteria. To his purified protein he then adds a branched 14-sugar oligosaccharide to the asparagine of the only Asn-X- Ser sequence found on GP1. Unfortunately, immune cells do not seem to recognize this synthesized glycoprotein. What's a likely explanation for this problem? O The oligosaccharide needs to be further modified before it's mature. O The oligosaccharide should've been added one sugar at a time. O The oligosaccharide needs a disulfide bond. O The oligosaccharideehould've been added to the serine instead of the asparagine.arrow_forwardYou are presented with Cytidine 5’ triphosphate and Thymidine 5’ triphosphate. Draw these phosphorylated structures as they would be connected in a polinucleotide in the order CpT / Show the individual phosphorylated structures first then show how they combine to form the polynuleotide. Show at any one of these structures where the glycosidic bond occursarrow_forward
- You are presented with Cytidine 5' triphosphate and Thymidine 5' triphosphate. Draw and upload to Efundi these phosphorylated structures as they would be connected in a polinucleotide in the order CpT / Show the individual phosphorylated structures first then show how they combine to form the polynuleotide. Show at any one of these structures where the glycosidic bond occursarrow_forwardOne common secondary structure found in proteins in the a-helix. On the diagram below is an a- helical region of the enzyme lysozyme. The sequence of this region is: NH Arg - Cys - Glu - Leu - Ala-Ala-Ala-Met-Lys. COO- 0 0 on the top 0 on the bottom R Is the N-terminus on the top or on the bottom of this diagram? Rarrow_forwarda. Suppose that you have the peptide Ala-Gly-Tyr-His-Leu and you treat it with FDNB and then 6M HCl. Draw the structures of all the products that you will have in solution (assume all reactions to go to completion).arrow_forward
- Most animals cannot use wool for nourishment. An exception is the clothes moth: it can derive nourishment from eating clothing made of natural fibers such as wool (which is rich in a-keratin). The digestive tract of the larvae of clothes moths provides a highly reducing environment. Why is this beneficial to the larvae? The reducing environment will reduce the disulfide linkages, making it easier to digest the proteins. The reducing environment will hydrolyze the peptide bonds. The reducing environment will interfere with the H-bonding between the tropocollagen molecules, making the type of protein found in most natural fibres easier to digest. The reducing environment will help to kill bacterial pathogens. The reducing environment will detoxify the carcinogenic dyes used in the textile industry.arrow_forwardMany enzymes are switched "on" by attachment of a phosphate group at a specific serine somewhere on the protein (phosphorylation). The basic reaction is: E + ATP2 Ep + ADP Po SERINE PHOSPHO SERINC (Note the "squiggles" before the backone amide and carbonyl indicate the polypeptide chain continues on either side of the serine). For phosphorylation to have this effect, there has to be some equilibrium between inactive and active forms conformations of the enzyme: [Eactive] [Einactive] Einactive 2 Eactive; K* The same basic equilibrium must exist for the phosphorylated protein: [Ep,active] [Ep,inactive] EP,inactive 2 Ep,active; Kp = (a) If phosphorylation increases the measured activity of the enzyme, is K* or K larger? Why? (b) Does the phosphorylation site need to be near the site where the enzyme binds its substrate (e.g. the reactant whose chemistry it catalyzes)? Why or why not?arrow_forwardPolypeptide folding is often mediated by other proteins called chaperones. Describe how a mutant chaperone protein might be responsible for a genetic disorder involving an enzyme.arrow_forward
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