Macromolecule Structures and Functions - Biochemistry

Aldotrioses are monosaccharides that contain both an aldehyde (an aldose) and three carbons (a triose). Knowing the definition of the word, and the breakdown of parts of the word, can help you recognize the molecule. The simplest aldotriose is glyceraldehyde.

A related concept involves ketotrioses, which are monosaccharides that contain both a ketone (a ketose) and three carbonds (a triose). Dihydroxyacetone is an example of a ketotriose.

Enantiomers are chiral molecules that are non-superimposable mirror images of each other. Diastereomers result when two or more stereoisomers of a compound have different configurations at one or more (but not all) of the equivalent stereocenters and are not mirror images of each other. An epimer is one of two stereoisomers that differ in configuration at only one stereocenter. An anomer is a type of epimer; it is one of two stereoisomers of a cyclic sugar that differs only in its configuration at the hemiacetal or acetal carbon (the anomeric carbon).

When it comes to metabolizing sugars, only reducing sugars are able to undergo breakdown. This is because reducing sugars are able to be converted from their closed chain form into their open chain form. It is only in the open chain form that sugars such as glucose can be metabolized.

Only reducing sugars can be converted into their open chain form. The reason for this is that the anomeric carbon for these sugars is not occupied. In their ring form, such sugars exist as hemiacetals that can readily and reversibly undergo chain opening. Additionally, some hemiketals can be converted into their open chain form, but they need to be able to tautomerize into their aldose form first.

Non-reducing sugars have their anomeric carbon tied up in a bond, and thus are locked in an acetal or ketal form. Consequently, they cannot convert into their open chain form, meaning that they cannot be metabolized.

In glycogen, glucose molecules are attached one after the other by alpha-1,4 linkages. However, in order to make glycogen more compact for storage, branch points are created to created links between many shorter glucose polysaccharides. These branch points connect glucose molecules by alpha-1,6 linkages.

Lots of alpha-1,4 linkages allow for longer chain lengths in carbohydrates like starch and glycogen. However, it is the amount of alpha-1,6 linkages that determine the number of branches - since glycogen has many more alpha-1,6 linkages than starch does, it has more branches. This allows for easy breakdown of glycogen into glucose in the liver should there not be enough glucose in the body to supply the body's demand for energy production. Recall that glycogen phosphorylase can only break terminal alpha-1,4 glycosidic bonds; hence, with more branches there are more terminal glucose molecules that are substrates for this catabolic enzyme.

Because glycogen is so heavily branched, it is able to pack more glucose units together in a small space, thus it is more compact and has a greater solubility. Moreover, the branching allows for glycogen enzymes to act more efficiently on the chains of glucose, and so both degradation and synthesis have increased rates.

For this question, we're told some background information about the liver's role in providing glucose homeostasis. We're told that when blood glucose levels are lowered, the liver is able to help restore glucose levels by keeping a large polysaccharide of glucose in store. In times of need, the liver can break this compound down to provide glucose to the bloodstream.

Even though the question doesn't explicitly tell us what the polysaccharide is, we should be able to infer that the compound in question is glycogen. Therefore, to answer the question, we need to know which kind of glycosidic bonds are found in glycogen.

First, let's recall that an individual glucose molecule is composed of six carbon atoms. In its ring form, glucose can exist as one of two epimers, depending on how its ring closes when transitioning from its straight chain form to its closed ring form. The anomeric carbon of the glucose molecule can be arranged in one of two ways when its ring closes. The anomeric carbon is the one that goes from being achiral to chiral as the ring closes. In the alpha configuration, the hydroxyl group attached to the anomeric carbon faces down, while in the beta configuration it faces up.

In addition to existing as either an alpha or a beta epimer, glucose also participates in glycosidic linkages using its first, fourth, and sixth carbon atom.

In glycogen, each individual glucose molecule is in the alpha configuration. Thus, we can rule out both answer choices that include beta. Moreover, the fourth carbon atom of each glucose molecule is attached to the first carbon atom (the anomeric carbon) in the next glucose molecule in the straight chain. To make branch points at various points along the straight chain, some of the glucose molecules have their sixth carbon atom attached to the anomeric carbon of other glucose molecules.

In conclusion, glycosidic bonds are responsible for branch points and glycosidic bonds are responsible for the straight chain.

lactose is made up of galactose and glucose and is bound via a beta 1,4 glycosidic bond.

the enzyme lactase cleaves this bond to break down the sugar lactose. Maltose is glucose- alpha 1,4- glucose, and sucrose is glucose- alpha, 1,2- fructose.

Glucose has five (not six) hydroxyl groups. Reducing sugars either have an aldehyde group or can form one through isomerism; sucrose doesn’t fit either description. Although there are more amylose molecules than amylopectin in starch, amylose is a minor component by mass; amylopectin makes up 70-80% of starch by mass. Polysaccharides are indeed joined in the union of two oses, which form glycosidic bonds.

Carbohydrate chains contain aldehyde or ketone functional groups, which are types of carboxyl groups. Remember the general formula for a carbohydrate is: since they are hydrates (water) of carbon.

To identify the answer choice that is NOT a polysaccharide, let's go delve into each answer choice a bit.

Cellulose is a polysaccharide and also a very important component of the cell wall of plants. It consists of many glucose sugars bound together via beta (1-4) linkages.

Peptidoglycan is also a polysaccharide and a very important component of the cell wall of bacteria. Its structure consists of alternating N-acetylglucosamine and N-acetylmuramic acid connected via beta (1-4) linkages. Both of these are modified sugar molecules.

Chitin is a polysaccharide that plays an important role in the structure of the exoskeleton of many fungi, arthropods, and insects. Its structure consists of a long chain of N-acetylglucosamine, which is a derivative of the sugar glucose.

Collagen is the only answer choice that does not represent a carbohydrate. Rather, collagen is a protein that plays a very important role in the extracellular matrix of various connective tissues found in animals.

Glycogen is a branched form of glucose that is the storage form in animals. It contains beta (1-4) and beta (1-6) linkages.

Cyclized monosaccharides can exist in two different stereoisomers that depend on the orientation of the hydroxyl group on the anomeric carbon. If this group is on the opposite side of the ring as the group, the sugar as the alpha anomer. The opposite orientation is referred to at the beta anomer. Cellulose is a polymer of beta glucose while starch is a polymer of alpha glucose, and the bonds between these glucose monomers differ depending on their anomer. This makes it so an enzyme can recognize one or the other but not both. Note that D-sugars are biologically relevant.

Amylose and amylopectin are the two components of starch. Approximately 80% of starch is composed of amylopectin. Both amylose and amylopectin are comprised of glucose units, and both have only D-glucose units. Amylose differs from amylopectin in that amylose is a single unbranched chain, but amylopectin branches. Therefore, amylopectin has both alpha 1,4 glycosidic linkages as well as alpha 1,6 glycosidic linkages.

A glycosaminoglycan (GAG) is a long, unbranched polysaccharide that consists of many repeating disaccharide units. Dermatan sulfate, keratan sulfate, hyaluronate, and heparin are all examples of glycosaminoglycans. Erythropoietin, however, is a glycoprotein hormone that stimulates production of red blood cells.

The induced fit model explains one method by which an enzyme's active site can accept some specific substrate. Initially, the active site might not be a perfect match for the substrate, however, when the substrate enters into the site, it can change the conformation of the enzyme just enough that it now fits perfectly and can be acted upon by the enzyme.

To answer this question, we need to have a general understanding about amino acid properties. For instance, at physiological pH, some amino acid side chains will carry a negative charge, some will carry a positive charge, and others will be neutral. Thus, we'll need to take note of which amino acid characteristics each position has, and then evaluate each answer choice to see if the new amino acid being substituted has different characteristics.

At position is arginine, which carries a positive charge. At position is valine, which has an aliphatic side chain that is neutral and relatively hydrophobic. At position is the amino acid glutamate, which is negatively charged due to the carboxyl group on its side chain. Finally, we have glycine at position , which contains a lonely hydrogen atom as its side chain.

Now that we have the characteristics of the amino acid residues in the enzyme, let's compare them to the substitutions listed in the answer choices.

Substituting an aspartate residue into position would mean replacing valine (neutral) with a positively charged amino acid. Hence, this would likely result in disruption of enzyme activity.

Substituting a tryptophan residue into position would replace glycine. In contrast to the extremely small side chain of glycine, the side chain of tryptophan is very large. This great size discrepancy could potentially lead to steric effects that could interfere with the binding of substrate to the enzyme.

Substitution of an asparagine residue into position would replace glutamate. Because glutamate is negatively charged, whereas asparagine is neutral, this substitution would likely interfere with enzyme activity.

Finally, let's consider the substitution of arginine at position with a lysine. In this case, a positively charged arginine would be replaced by another positively charged amino acid, lysine. Because of the similarity between these two amino acids, this substitution would be the least likely to cause a disruption in the enzyme's activity.

In a globular protein, the amino acid chain can twist in a way that polar groups lie at the protein's surface. This allows the protein to interact with water and enhances the protein's solubility in water. This does not occur in fibrous proteins, so fibrous proteins are insoluble in water.

Actin chain growth occurs at the (+) end of the chain, and nucleotide hydrolysis promotes dissociation of actin chains. 2 microfilaments of G-actin monomers make 1 filament of F-actin. However, actin filament assembly is powered by ATP, not GTP.

An O-linked glycoprotein is a protein that has a sugar attached to it. It is called O-linked because the sugar is attached to an oxygen atom on either a threonine residue or a serine residue within the protein.

A peptide bond is formed via the condensation of one amino acid's alpha-carboxy group with the alpha-amino group of another amino acid. Thus, the joining together of two amino acids results in the loss of one water molecule. Likewise, joining three amino acids together results in the loss of two water molecules. Following this pattern, we can conclude that the number of water molecules lost is equal to the number of amino acids joined together, minus 1. Therefore, the joining together of 100 amino acids results in the loss of 99 water molecules.