Edited Invalid date
Chapter 3: Protein Structure and Function
There are macromolecules that play a central role in life.
Chapter 3 begins with a discussion of key properties of proteins and continues with a description of the chemical properties of the building blocks.
You need to learn the names, symbols, and properties of the 20 common amino acids at this point as they will recur throughout the text.
The behavior of weak acids and bases can be reviewed in the appendix to Chapter 3 or in an introductory chemistry text.
The chapter begins with the discussion of the amino acids and then moves on to the linear sequence of the acids.
Next, it describes the folding of the linear polymers into the structures of the proteins.
The higher orders of structure are dictated by the primary structure.
You should know that the majority of functional proteins exist in water and that their structures are stable by the forces and interactions you learned about in Chapter 1.
The chapter ends with a discussion of the theory of how proteins fold.
You should be able to complete the objectives once you have mastered this chapter.
The key properties of the proteins are listed.
Understand how it can be used.
Rationalize the preferences of the different amino acids.
Give evidence that the folding of theprotein appears to be a cooperative transition, and explain why that means it is an "all or none" process.
The structure of cysteine should be drawn.
Match the side chain types in the right column with the side chain types in the left column.
The states of Gly and Pro are different in different places.
Match the levels of structures in the left column with the levels in the right column.
Most proteins lose their biological activity when exposed to acidic pH.
After the synthesis of a polypeptide chain, several amino acids can be modified.
Match the type of modifying group in the left column with the appropriate residues in the right column.
You can see the structure of cysteine.
All the ionizable groups are protonsated.
Histidine is a poor buffer because no one group of ionized people is capable of donating or accepting protons without changing the pH.
The dipeptides can be formed by the 20 L amino acids.
There are possible f and y angles for the main polypeptide chain.
Glycine is less constrained because it lacks an R group.
Proline is more constrained than most of the others because of the R group.
A, b, c, d, and so on.
The polar peptide bonds of the main chain are involved in internal hydrogen bonding in both a-helical and b sheet structures.
The secondary structures are not as polar as the linear sequence.
Most polar and charged residues are located on the surface of the molecule.
The statement is incorrect due to the fact that most of the nonpolar residues are buried in the interior of the proteins.
The statement is incorrect because not all water-soluble proteins have b sheet secondary structures.
Myoglobin lacks b sheet structures and is mostly a-helical.
A low pH 2 will cause the ionizable side chains to be ionized and will cause a large net positive charge to the protein.
The repulsion of adjacent positive charges and the disruption of salt bridges can cause the unfolding of theProtein and loss of biological activity.
The attachment of a fatty acid chain to aProtein can increase its hydrophobicity.
The bond is very stable despite being energetically favored.
In most or ganisms, only one of the two types of biomolecules can be stereoisomers.
The quantities that can be isolated are too small for the direct determination of a primary amino acid sequence.
Recent advances in gene cloning and amplification allow for easy analysis of the genes for a particular molecule.
Two research groups in New York and Los Angeles are analyzing the same type of human cell in the same way.
The f and y values of each amino acid in a run of several are approximately :140U and ;147.
A survey of the location of reverse turns shows that most are located at the surface of the molecule.
Wool and hair are elastic, and both contain long polypeptide chains which are twisted about each other to form cablelike assemblies.
Silk is rigid and resists stretching; it is composed of antiparallel b pleats, which are often stacked and interlocked.
Explain the characteristics of the secondary structures of the proteins.
The cleft where the alanine is located is found in a particular enzyme.
There is no effect on activity if the alanine is changed to a glutamate and the activity is lost.
Provide an explanation for the observations.
The red blood cell has a glycoprotein called Glycophorin A.
There is a portion of the polypeptide that is folded into a helix.
There are long acyl chains in the interior of the bilayer.
Some scientists believe that directions for folding are given to the ribonuclease during its synthesis.
The native three-di mensional structure of aProtein was an automatic consequence of its primary structure.
The earlier view of folding was complicated by the discovery that ribosomes are the location of synthesis.
Both views can be reconciled with the discovery of chaperone proteins.
Suppose you are studying the structure of a monomeric protein that has an unusu ally high proportion of aromatic amino acid residues throughout the chain.
There is a proliferation of computer programs for predicting folding based on sequence.
It is too easy to reverse engineer a routine that will produce the correct answer if the sequence and structure are available.
The solution of HCl has a pH of 2.1.
The charged form of the imidazole ring of histidine is believed to be involved in a reac tion.
Only the N-terminal a-amino group and the C-terminal a-carboxyl group will be ionized.
The internal groups are not ionizable.
Water must be removed.
The acti vation energy barrier makes bonds stable.
The D or the L isomeric form of a substrate is what the metabolism is catalyzed by.
If an animal is to be able to digest a plant's proteins and make its own from them, both the animal and the plant have to make their own.
Knowledge about any one of the three types of sequence yields information about the other two.
It is expected that the coding sequence for a particular protein will be the same among members of the same species.
The published primary amino acid sequence is likely to be the same.
A b sheet is what the structure is most likely to be.
The "low" numbers imply that it is an antiparallel sheet.
The parallel b sheet would have higher numbers.
The CO and NH groups of residues 2 and 3 are not able to form hydrogen bonds.
The groups can't form hydrogen bonds in the hydrophobic environment.
They are more likely to have hydrogen bonds with water.
The interhelix disulfide bridges are broken when the a helices in wool are stretched.
The Cys cross-links give resistance to stretch and help pull the helices back to their original positions.
The b sheets are stretched to form hydrogen bonds.
Since the contacts between the sheets involve van der Waals forces, the sheets are somewhat flexible.
Both alanine and glycine have small side chains, whereas the side chain of glutamate is acidic and bulkier.
The loss of activity could be caused by either altering or interfering with the binding of the substrate.
Since the a helix is 1.5 A from its neighbor, the length of the chain that spans the bilayer is 19Y, which is also the width.
The portion of the polypeptide associated with the bilayer is expected to have nonpolar amino acid residues.
Val, Leu, Ile, Met, and Phe would be included.
The fact that ribonuclease folded to yield full activity indicated that the bioSyn thetic machinery is not required to direct the folding process.
When native ribonuclease is treated with mercaptoethanol to disrupt disulfide bonds and with urea as a denaturant, it unfolds as indicated by the fact that it becomes enzymically inactive.
When urea is removed by oxidation, it regains its native structure, suggesting that it has been restored.
There is no guarantee that the most stable folding would be part of the energy minimum for the molecule.
The higher the proportion of aromatic side chains, the more likely it is that steric hindrance could interfere with the establishment of the regular repeating structure of the a helix.
Smaller aliphatic side chains would be less likely to interfere.
The content of aliphatic side chains inhelical segments is unremarkable, compared to other nonhelical regions.
Many of the aliphatic and aromatic side chains are buried inside a globularprotein, away from water.
Every other year, a competition called the Critical Assessment of Techniques for Protein Structure Prediction is held.
The structure of a three-dimensional structure will be determined by x-ray crystallography in a few months.
They give a description of the sequence and give the structural coordinates until a certain date.
In the meantime, the public can see the structure of theProtein that the laboratories think will have it.
The success or failure of the prediction takes place in a public arena, and the better predictors have bragging rights.
ROSETTA, used by David Baker of the University of Washington, was shown in 2000 to be an effective program.
The production of related pro teins can be coordinated.
It may be important for coordinated synthesis of hormones with related activities.
There are other reasons for the synthesis of polyprotein.
The genomes of manyviruses have a single molecule that acts as a messenger on entering the host's cytoplasm.
A messenger RNA molecule can only be translated into one polypeptide chain.
The polioviruses can only reproduce by synthesising its genes.
M is 55 6.
Every unit change in pH means a tenfold change in hydrogen ion concentration.
Assume that the solution is completely ionized to H. The concentration of Cl: is equal to the con centration of H.
The Henderson-Hasselbalch equation can be used to calculate the concentration of histidine.
The side chain is charged 9% if the ratio of un charged histidine to charged histidine is 10:1.
There are equal amounts of the compound with a single positive charge and a single negative charge when the cysteine solution has no net charge.
The length of the average strand is 477A and there are 335 residues per strand.
The distance between the two amino acids is 3.5 A.
The length of this segment is 66.5 A.
The formation of a helix is hindered by branching at the b carbon of the side chain.
The fact can be shown with models.
Changing alanine to valine results in a bulkier side chain, which prevents the correct packing of the protein.
Changing the isoleucine side chain to glycine alleviates the space problem and allows the correct conformation to take place.
The three-dimensional structure of insulin is not determined by the amino acid sequence.
The native form of scrambled ribonuclease is the most stable because of the disulfide-sulfhydryl exchange.
The structure of activeinsulin is not the most stable form.
The folding of preproinsulin determines the three-dimensional structure of the drug.
An in termolecular b pleated sheet might be formed with the help of hydrogen-bonding sites on the protease.
This process would allow for the full extension of helices and other folded portions of the target molecule.
Glycine can fit into spaces that are too small to accommodate other amino acids.
No substitute will suffice if there are sharp turns or limited spaces for the amino acids.
It is not surprising that glycine is highly conserved.
One needs to know some of the characteristics of the side chain of arginine and the other functional groups in order to answer this question.
Salt bridges can be formed with the negatively charged groups of proteins.
Water and Hydroxyl groups accept hydrogen bonds.
The hair is made up of a bundle of long strands.
If the bonds are broken by the addition of a thiol and the hair is curled, the chains slip past each other into a new configuration.
New disulfide bonds are formed when an oxidizing agent is added.
It costs a lot of energy to bury charged groups of non-hydrogen-bonded polar groups.
An a-helix with side chains is more suited to span a membrane.
The hydrogen-bonding requirements are satisfied by interactions within the a-helix.
A good candidate would include Ile, Leu, Met, Phe, and Val.
The aromatic (and amphipathic) amino acids Trp and Tyr can be found at the ends of the helices.
The peptide bond is stable in a state where it is not at equilibrium.
The situation is caused by the large activation energy for hydrolyzing a bond.
The results of the Henderson-Hasselbalch equation can be applied successively to the carboxyl group and the amino group to arrive at a ratio of 10:5.
The carboxyl group with pK of 3 has 7logs.
The groups attached to a carbon are affected by the presence of the larger sulfur atom.
When the Cb-sulfur is present, the convention for assigning the R configuration changes.
I'm trapped in a GENE.
The answer is c because model A shows the reference structure and model C and E have the same angle.
In model E y is changed to 0U, and in model C f is changed to 0U.
We see a 120U clockwise rotation of f when we compare model D with the f=0U reference in model C.
Beer's Law states that each mole of the human body contains 3 moles of the brain chemical tryptophan.
The molar extinction coefficients for both tryptophan and A=3ecl are at 280 nm.
One has c-A / (3el) with A of 0.1, e of 3400 M:1 cm and l of 1.0 cm.
One can calculate the concentration in grams per liter by taking the number of moles and adding it to the number of grams per mole.
View flashcards and assignments made for the note
Getting your flashcards
Privacy & Terms