Which types of hydrogen bonds are known to exist in RNA secondary structures?

Which types of hydrogen bonds are known to exist in RNA secondary structures?

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I'm programming an implementation of an algorithm for pattern matching in RNA structures. The algorithm assumes the following types of base pairs:

  • Plain: No base pairs (just the primary structure of the RNA)
  • Single: Each base can be connected to at most one other base
  • Multiple: Each base can be maximally connected to a number of other bases (where 'number' can be constant or infinite)

Although there is no problem with the implementation of the above types of base pairs (apart from the probable high computation complexity), I wonder if all of them were observed in real RNAs?

In particular, I'm interested in the following cases:

  • A base pair between two adjacent bases i,j.
  • One base (i) which connected to two different bases (j,k) (i is not adjacent to j or/and k).
  • One base (i) which connected to more than two different bases, all of them not adjacent to i.

Can these exist? Some of them? If they can, how common is their existence?

A base pair between two adjacent bases i,j.

It is not possible. To form an intramolecular hydrogen bond, the RNA has to bend. The persistence length i.e. the minimum length of the chain required for bending, is around 4 bases for ssRNA (I am not very sure about this number. At this moment I am not able to locate the exact reference. I remember this number from the back of my head which I heard somewhere).

One base (i) which connected to two different bases (j,k) (i is not adjacent to j or/and k).

It is definitely possible for one pair after considering the persistence length. If a single base is bound to more than two bases then atleast one interaction should have to form a non Watson-Crick type bonding, as indicated by Mad Scientist.

One base (i) which connected to more than two different bases, all of them not adjacent to i

Very less likely. There are no known reports. Even if it happens, the bond will be weak because of limitation in the number of H-bond donor/acceptor sites in the bases.

By far the most common type of base pair is the Watson-Crick base pair in an RNA helix. Those are comparably easy to predict, e.g. Mfold and the Vienna RNA package can do this.

Base triples, three nucleobases that form hydrogen bonds to each other are not uncommon in RNAs with a complex tertiary structure. There is even a database of RNA triples, though this one also contains triples where not all three bases make contact with each other.

There are also quadruplexes, but there each base only makes contact with two other bases. I don't know if there are any structures where one base makes contact with more than two other ones.

Nitrogenous Bases

Chemical structures of the five nitrogenous bases are shown below. The red N atom in each molecule is the point of attachment for a sugar molecule (ribose or deoxyribose).

Adenine and guanine are purine bases found in both DNA and RNA.
Cytosine is a pyrimidine base found in both DNA and RNA.
Thymine and uracil are pyrimidine bases found in only DNA or RNA, respectively.

These five nitrogenous bases are all planar molecules, meaning that they are fairly flat and rigid.

Examine the structure of one nitrogenous base molecule by clicking on the button below (wait a few seconds for it to load in the space at right). You may click and drag the molecule to rotate it in three dimensions.

The green bond represents the point of attachment for a sugar (ribose or deoxyribose).

What is the nitrogenous base displayed in the computer model?

Molecular Cell Biology. 4th edition.

Carbohydrates illustrate the importance of subtle differences in covalent bonds in generating molecules with different biological activities. However, several types of noncovalent bonds are critical in maintaining the three-dimensional structures of large molecules such as proteins and nucleic acids (see Figure 2-1b). Noncovalent bonds also enable one large molecule to bind specifically but transiently to another, making them the basis of many dynamic biological processes.

The energy released in the formation of noncovalent bonds is only 1 –𠁕 kcal/mol, much less than the bond energies of single covalent bonds (see Table 2-1). Because the average kinetic energy of molecules at room temperature (25 ଌ) is about 0.6 kcal/mol, many molecules will have enough energy to break noncovalent bonds. Indeed, these weak bonds sometimes are referred to as interactions rather than bonds. Although noncovalent bonds are weak and have a transient existence at physiological temperatures (25 –  37 ଌ), multiple noncovalent bonds often act together to produce highly stable and specific associations between different parts of a large molecule or between different macromolecules (Figure 2-11). In this section we consider the four main types of noncovalent bonds and discuss their role in stabilizing the structure of biomembranes.

Figure 2-11

Multiple weak bonds stabilize specific associations between large molecules. (Left) In this hypothetical complex, seven noncovalent bonds bind the two protein molecules A and B together, forming a stable complex. (Right) Because only four noncovalent (more. )

1.4 Properties of Biological Macromolecules Overview

This overview covers section 1.4 of the AP Biology Curriculum – Properties of Biological Macromolecules.

Let’s start with arguably the most important biological macromolecule: Nucleic acids. To fully understand how nucleic acids work, we need to look at their structure. First, let’s take a look at the sugar-phosphate backbone of a nucleic acid.

At the center of every nucleic acid is the sugar-phosphate backbone. The phosphate group forms phosphoric acid in water. This phosphate group can bond to the sugar molecule on the next nucleic acid, creating a long chain. No matter how long this sugar-phosphate backbone gets, there will always be a phosphate group exposed on one end and a sugar molecule exposed on the other. Therefore, we call both a single nucleotide and many nucleotides connected together a “nucleic acid.”

The main difference between DNA and RNA lies in the sugar molecule that is used to create the sugar-phosphate backbone. DNA uses deoxyribose, seen here. RNA uses ribose – the same sugar with one extra oxygen atom. This tiny difference creates some of the functional differences between DNA and RNA within cells.

The part of a nucleotide that is most important to carrying information is the nucleotide base. The base attached to this structure is cytosine, one of several bases that can be attached to a nucleotide. Let’s see exactly how these nitrogenous bases work.

There are 5 nitrogenous bases used in nature to create DNA and RNA, which are separated into two groups based on their structure. The purines are based on a double-ring structure, whereas the pyrimidines are based on a single-ring structure. Adenine, Guanine, Thymine, and Cytosine are used to create DNA molecules. Uracil is used in RNA, in place of Thymine.

More importantly, the nitrogenous bases create the double-helix structure of DNA through their ability to form hydrogen bonds. Each purine has a corresponding pyrimidine that it can form hydrogen bonds with. You can remember which nitrogenous bases can form hydrogen bonds using a simple mnemonic device. The tall letters (A + T) can form hydrogen bonds, and the fat letters (C + G) can form hydrogen bonds. This will be very important to remember when we start to learn how DNA is synthesized and how errors in the DNA code are corrected.

DNA stores information through a slightly complex mechanism. DNA is stored in the nucleus as a double helix. This allows it to stay protected from damage. The double-helix also allows for repair proteins to easily find errors. Most errors create a small bump in the DNA, due to the lack of hydrogen bonds between the two strands. To extract the information needed to create new proteins, the exact order of nucleotides must first be copied from DNA into a new RNA molecule within the nucleus. This is called transcription.

RNA is not as stable as DNA, and is more prone to errors. However, RNA molecules can carry the information to where it is needed – like a messenger. This messenger RNA molecule carries the nucleotide sequence out of the nucleus, where a ribosome can attach to it. The ribosome then creates a new protein molecule by matching transfer RNA molecules to every 3-nucleotide sequence, known as a “codon”. This process, called translation, is how the information stored in DNA becomes an actual cellular product and allows the cell to function.

Now that we know how DNA stores the information to build proteins, let’s take a look at proteins themselves. Proteins are simply large strings of amino acids that fold into specific shapes. Each protein serves a different function, made possible by its 3-dimensional shape and the amino acids it is made of.

Amino acids – also called peptides – are bonded together by peptide bonds. These bonds form through a dehydration reaction between a carboxyl group and an amino group on each amino acid. This also ensures that each protein molecule has directionality. One one side is the carboxyl terminus, while the other side of the molecule has an amino terminus. Make sure you understand the difference because questions on the AP test can reference these different sides.

The structures that make each amino acid different are known as R-groups or side-chains. These groups are what gives each amino acid its unique functionality. In fact, though there are 20+ amino acids used in nature, there are only 7 different groups that these molecules can be classified as. While the structure of each amino acid is slightly different, many amino acids bring similar properties to the polypeptides they are a part of.

For instance, several amino acids have charged R-groups. This helps create a hydrophilic portion of the polypeptide that can easily interact with water and other polar molecules. Other amino acids contain sulfur, which is able to form sulfur crosslinks with other sulfur-containing peptides. This can help hold multiple polypeptides together in a large quaternary structure.

The active site is where the protein will actually carry out its function. In order to fit a substrate just right and catalyze a reaction, the active site of the protein must have the right physical and chemical properties. So, not only does the active site need to have the right R-groups exposed, but the protein must also have the right sequence of amino acids to fold into the proper shape.

Likewise, this protein must also have some hydrophobic regions where it needs to bind to the cell membrane. If hydrophilic amino acids were used in place of hydrophobic amino acids, this protein could not stick within the cell membrane and would not be functional. Since proteins serve roles as enzymes, immune responders, receptors, methods of movement, and as structural molecules, there is a nearly infinite number of amino acid arrangements.
Carbohydrates most commonly serve roles as fuel and building materials for a cell. The simplest carbohydrates are hydrocarbon chains of 5 or 6 carbons that often have a ring-like structure. Glucose, for instance, serves as the main fuel molecule for cells. However, as you connect more and more carbohydrate monomers, you can create substances with many different properties.

The exact structure of large polysaccharides helps determine their function. Linear polymers are most often found in structural molecules like cellulose. These fibers – much like the smaller threads in a large rope – can intertwine to create a much stronger material. Some structural carbohydrates even have cross-links between the fibers, adding another layer of strength to a molecule.

By contrast, storage polysaccharides most often have a branched structure. Unlike a linear structure, this allows a cell to store as much energy in as small of a space as possible. Starch molecules – such as amylose found in potatoes – are essentially huge branching structures that fill cells with energy. Humans and animals use the polysaccharide glycogen for a similar purpose. The cell can easily start hydrating the bonds between individual monomers to fill the cell with glucose – which can then be used to power a variety of other reactions.

The last category of macromolecules that we will look at is lipids. There are three types of lipids that are most important to life: fats (triglycerides), phospholipids, and steroids. Some people consider waxes their own category, though they have a structure very similar to triglycerides. Let’s take a look at each of these groups.

Triglycerides are simply fatty acid molecules bound into a larger molecule with glycerol – a three-carbon alcohol. Fatty acids come in two forms: saturated and unsaturated. Palmitic acid is an example of a saturated fatty acid. Every carbon in the chain is bound to at least 2 hydrogens, leaving no room for double bonds between carbon atoms. Structurally, this makes saturated fats very linear. Therefore, you can pack many saturated fatty acids into a very tight space. Because of this structure, saturated fatty acids are usually solid at room temperature because the molecules squeeze tightly together as they lose thermal energy.

By contrast, an unsaturated fatty acid has double bonds between at least 2 carbon atoms in the chain. Double bonds are rigid. This means that lots of fatty acids cannot pack tightly together if they are unsaturated – even if the temperature is not particularly warm. Olive oil is a good example of an unsaturated fatty acid.

To create a triglyceride, three fatty acids bind to a single glycerol molecule. Though lipids are not “true polymers” in the sense that they are linear chains of the same monomers, they are still created through dehydration reactions. The hydroxyl groups on glycerol react with the carboxyl head groups of each fatty acid. A water molecule is lost and an ester bond is formed. There are many triglycerides found in nature, with both saturated and unsaturated fatty acids in their structure. This gives rise to many different types of fat found in different organisms.
Phospholipids are different structurally – compared to triglycerides – and they also serve a much different purpose within organisms. Phospholipids have a hydrophilic head and a hydrophobic tail. When many phospholipids congregate together, the head groups interact with water while the tail groups tend to orient toward each other. This is how the lipid bilayer of all cells is created. Let’s look closer at the structure of a phospholipid.

In the hydrophobic tail are long hydrocarbon chains. The tail sections can contain saturated or unsaturated fatty acids, depending on the organisms. In general, organisms that live in very hot environments tend to have more saturated fatty acids whereas cells that must exist at very low temperatures tend to have more unsaturated fatty acids. Since unsaturated fatty acids tend to remain liquid at low temperatures, this creates a cell membrane that is still fluid and functional in the cold. Each organism must maintain the right balance of fatty acid tails to ensure its cells have functional membranes.

The polar head groups of phospholipids have both phosphate groups and nitrogen – both of which increase the head’s hydrophilic tendencies. This ensures that the molecule’s head is always oriented towards water – whether that is the cytosol of the cell or the external environment.


In this study, we characterized a data set of 91 protein–RNA complexes to identify novel mechanisms underlying protein–RNA interactions. Our data indicate that both the surface shape of the protein and the secondary structure of the RNA molecule are important in determining the binding specificity of a given protein and RNA molecule. We observed that a dented protein surface is significantly more likely to interact with unpaired nucleotides, and the hydrogen bonds at this interface are prominent between the protein backbone and RNA bases. Indeed, previous studies have shown that hydrogen bonds frequently occur between the protein backbone and RNA bases ( 6 , 7 ) and that a protein cavity (i.e. a dented surface) prefers nucleotide bases at the interface ( 9 ). Gupta and Gribskov ( 25 ) extensively analyzed the base pairing property in RNP region (i.e. interface) using different data set and reported the preference of unpaired nucleotides in this region. Thus, our results are consistent with these previous reports and also show that a dented protein surface can distinguish unpaired nucleotides from paired nucleotides through hydrogen bonds that form between the protein backbone and RNA bases at the interface. Consistent with previous reports ( 6–8 ), we also observed that positively charged amino acids often form electrostatic interactions with the phosphate groups of RNA. Interestingly, this type of interaction was more often observed on proteins with a protruded surface. Collectively, these data suggest that dented and protruded protein surfaces employ different recognition mechanisms for paired versus unpaired RNA nucleotides. We further hypothesize that protruded protein surface makes an initial contact with the RNA molecule through electrostatic interactions, and a dented surface determines the binding specificity through the hydrogen bonds that form with unpaired nucleotides.

A loop region is one of the major structural features of RNA and is frequently used to form an interaction with various RNA binding proteins ( 19 ). Nucleotide bases in RNA loops exhibit unique hydrogen-bonding patterns with proteins, and these patterns are key determinants of binding specificity ( 6 , 26 ). In this study, we found that aspartic acids interacted more frequently with the RNA loops in which the nucleotide bases had flipped out to form hydrogen bonds with the protein. Aspartic acids are generally disfavored at protein–RNA interfaces due to electrostatic repulsion between negatively charged side chains and phosphate groups ( 6–8 ). However, aspartic acids are also known to form specific pseudo pairs with the nucleotide bases by using both their side- and main-chain atoms ( 27 ). Based on our results and previous reports, we speculate that aspartic acids are necessary for base-flipping, most likely to keep phosphate groups away from an interface and to form some specific interactions with the flipped bases.

Protein–RNA interactions are controlled by various factors, such as the composition of the amino acids and nucleotides, the shape of the macromolecules and higher order structures. Our study highlights the roles that are played by the protein surface and the secondary structure of the RNA molecule in protein–RNA interactions and also suggests a possible role of aspartic acid in RNA loop recognition. However, there are many important issues that need to be addressed for understanding the mechanism of protein–RNA interactions. For example, protein surface shapes of RNA interacting proteins should be compared with the other proteins, such as those that bind to DNA or ligands, to identify unique characteristics of such proteins. A number of prediction algorithms for protein–RNA interactions are available ( 28–34 ), and the inclusion of features such as the shape of the protein surface and the secondary structure of the RNA molecule will greatly improve the efficiency and accuracy of these algorithms.

Similarities Between Primary Secondary Tertiary Structure of Protein

  • Primary, secondary, and tertiary structure are three, structural arrangements of proteins.
  • The basic unit of all of the structures is the amino acid sequence, which is the primary structure of protein.
  • Secondary structure of protein is formed from its primary structure, which in turns form the tertiary structure.
  • Each type of structure has a unique role in the cell.


Ali on January 01, 2019:

Nigga on March 26, 2018:

The only bond i love is the one between me and fortnite

gotta leave earth on August 30, 2017:

What is bonding between lovers called? :)

sachin on September 30, 2016:

swathi on September 10, 2016:

this notes or article is really easy to understand the concepts, now am getting to know the concepts of bonding , and thier types, strength of bonding ,which bonding is more stronger and all . so thanks to giving the article like this.

ISHAAN on September 09, 2016:

the best is metallic bond as it makes the substance way more strong!

PHD on July 28, 2016:

Van der Waals bonding is weeeeeaaaaaaaakkk, hydrogen bonding is the best.

rrr on July 28, 2016:

manjit on February 03, 2016:

Plz give a proper example of bondings

Yashwanth on January 02, 2016:

its good and easy to learn

fekade begosew on March 28, 2015:

It is best continue in the such way

Xavier Luther on February 02, 2015:

Why does no one like Van der Waals bonding? THEY ARE THE BEST.

wangno on December 08, 2014:

As per strenght is concerned, i think ionic bond is the strongest not metallic followed by covalent then metallic among primary bonds..

ria-majumdar (author) from Manipal on February 18, 2012:

Thanks for the comment. Your points are correct where strength is concerned.

taheruddin from Khulna Bangladesh on February 18, 2012:

vote up, awesome, very nice article to learn about bond. Thanks, Ms. Ria Majumdar.

I can not but say you can add the issue of strength of bond. As I know metallic is the strongest, then ionic, then covalent. This strength also has relation with its state solid, liquid and gas. Stronger bond make the substance harder means solid and weaker bond let it be liquid or gas.

Donors and Acceptors

In order for a hydrogen bond to occur there must be both a hydrogen donor and an acceptor present. The donor in a hydrogen bond is usually a strongly electronegative atom such as N, O, or F that is covalently bonded to a hydrogen bond.

The hydrogen acceptor is an electronegative atom of a neighboring molecule or ion that contains a lone pair that participates in the hydrogen bond.

Why does a hydrogen bond occur?

Since the hydrogen donor (N, O, or F) is strongly electronegative, it pulls the covalently bonded electron pair closer to its nucleus, and away from the hydrogen atom. The hydrogen atom is then left with a partial positive charge, creating a dipole-dipole attraction between the hydrogen atom bonded to the donor and the lone electron pair of the acceptor. This results in a hydrogen bond.(see Interactions Between Molecules With Permanent Dipoles)

NCERT Solutions for Class 12 Chemistry Chapter 14 Biomolecules

NCERT Solutions for Class 12 Chemistry Chapter 14 Biomolecules are been solved by expert teachers of All the solutions given in this page are solved based on CBSE Syllabus and NCERT guidelines.

INTEXT Questions

Question 1.
Glucose or sucrose are soluble in water, but cyclohexane or benzene (simple six membered ring compounds) are insoluble in water. Explain.
Glucose or sucrose contain several hydroxyl groups in their molecules which form hydrogen bonding with water molecules due to which they dissolve in water. On the other hand compounds like benzene or cyclohexane cannot form hydrogen bonds with water molecules, so they are insoluble in water.

Question 2.
What products are expected when lactose is hydrolysed ?
Lactose (C12H22O11) on hydrolysis with dilute acid yields an equimolar mixture of D-glucose and D-galactose.

Question 3.
How do you explain the absence of aldehyde group in pentaacetate of glucose ?
The cyclic hemiacetal form of glucose contains an OH group at C-1 which gets hydrolysed in the aqueous solution to produce the open chain aldehydic form which then reacts with NH2OH to form the corresponding oxime. Therefore, glucose contains an aldehydic group. On the other hand, when glucose is reacted with acetic anhydride, the OH group at C-1, along with the four other OH groups at C-2, C-3, C-4 and C-6 form a pentaacetate. As the pentaacetate of glucose does not contain a free OH group at C-1, it cannot get hydrolysed in aqueous solution to produce the open chain aldehydic form and thus glucose pentaacetate does not react with NH2OH to form glucose oxime. Hence, glucose pentaacetate does not contain the aldehdye group.

Question 4.
The melting points and solubility in water of amino acids are higher than those of the corresponding haloacids. Explain.
The amino acids exist as zwitter ions, H3N + OHR – COO – . Because of this dipolar salt like character they have strong dipole- dipole attractions. So, their melting points are higher than halo acids which do not have sail like character. Moreover, due to this salt like character, they interact strongly with H2O. Thus, solubility of amino acids in water is higher than that of the corresponding halo acids which do not have salt like character.

Question 5.
Where does the water present in the egg go after boiling the egg.
The boiling of an egg is a common example of denaturation of proteins present in the white portion of an egg.

The albumin present in the white of an egg gets coagulated when the egg is boiled hard. The soluble globular protein present in it is denatured resulting in the formation of insoluble fibrous protein.

Question 6.
Why vitamin C cannot be stored in our body?
Vitamin C is a water-soluble vitamin. Water-soluble vitamins when supplied regularly in the diet cannot be stored in out body because they are readily excreted in urine.

Question 7.
What products would be formed when a nucleotide from DNA containing thymine is hydrolysed ?
When a nucleotide from DNA containing thymine is completely hydrolysed, the products obtained are :

  1. 2-deoxy-D(-)ribose.
  2. two pyrimidine i.e., guanine (G) and adenine (A).
  3. two purines, i.e., thymine (T) and cytosine (C) and
  4. phosphoric acid.

Question 8.
When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained ? What does this fact suggest about the structure of RNA ?
A DNA molecule has two strands in which the four complementary bases pair each other, viz. cytosine (C) always pairs with guanine (G) while thymine (T) always pairs with adenine (A). Therefore, when a DNA molecule is hydrolysed the molar amount of cytosine is always equal to that of guanine and that of adenine is always equal to that of thymine RNA also contains four bases, the first three are same as in DNA but the fourth one is uracil (U).

As in RNA there is no relationship between the quantities of four bases (C, G, A and U) obtained, therefore, the base-pairing principle, viz., (A) pairs with (U) and (C) pairs with (G) is not followed. So, unlike DNA, RNA has a single strand.

NCERT Exercises

Question 1.
What are monosaccharides?
A carbohydrate that cannot be hydrolysed further to give simpler unit of polyhydroxy aldehyde or ketone is called a monosaccharide. With a few exceptions they have general formula, C,,H2„O„. About 20 monosaccharides are known to occur in nature. Some common examples are glucose, fructose, ribose, etc.

Question 2.
What are reducing sugars?
All those carbohydrates which reduce Fehling’s solution and Tollen’s reagent are referred to as reducing sugars. All monosaccharides whether aldose or ketose are reducing sugars.

Question 3.
Write two main functions of carbohydrates in plants.
Two main functions of carbohydrates are

  1. Cell wall of bacteria and plants is made up of a polysaccharide, cellulose.
  2. Starch is the major food reserve material in plants.

Question 4.
Classify the following into monosaccharides and disaccharides. Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.
Monosaccharides : Ribose, 2-deoxyribose, galactose and fructose
Disaccharides : Maltose and Lactose

Question 5.
What do you understand by the term glycosidic linkage?
Disaccharides on hydrolysis with dilute acids or enzymes yield two molecules of either the same or different monosaccharides. The two monosaccharides are joined together by an oxide linkage formed by the loss of a water molecule. Such a linkage between two monosaccharides units through oxygen atom is called glycosidic linkage.

Question 6.
What is glycogen? How is it different from starch ?
The carbohydrates are stored in animal body as glycogen. It is also known as animal starch because its structure is similar to amylopectin. It is present in liver, muscles and brain. When the body needs glucose, enzymes break the glycogen down to glucose. Glycogen is also found in yeast and fungi.

Starch is the main storage polysaccharide of plants. It is the most important dietary source for human being. High content of starch is found in cereals, roots, tubers and some vegetables. It is a polymer of two components-amylose (15-20%) which is water soluble and amylopectin(80-85%) which is water insoluble.

Question 7.
What are the hydrolysis products of (i) sucrose and (ii) lactose ?

  1. Sucrose on hydrolysis gives one unit of glucose and one unit of fructose.
  2. Lactose on hydrolysis with dilute acids yields an equimolar mixture of D-glucose and D-galactose.

Question 8.
What is the basic structural difference between starch and cellulose ?
The basic structural difference between starch and cellulose is of linkage between the glucose units. In starch, there is a-D-glycosidic linkage. Both the components of starch-amylose and amylopectine are polymer of α-D-glucose. On the other hand, cellulose is a linear polymer of β-D-glucose in which C1 of one glucose unit is connected to C4 of the other through β-D-glycosidic linkage.

Question 9.
What happens when D-glucose is treated with the following reagents?


Question 10.
Enumerate the reactions of D-glucose which cannot be explained by its open chain structure.
The open chain structure of D-glucose OHC – (CHOH)4 – CH2OH fails to explain the following reactions :

(i) Though it contains the aldehyde (-CHO) group, glucose does not give

2,4-DNP test, Schiff’s test and it does not form the hydrogen sulphite addition product with NaHSO3.

(ii) The pentaacetate of glucose does not T react with hydroxylamine (NH2OH) to form ’ the oxime indicating the absence of free -CHO group.

(iii) The formation of two anomeric methyl glycosides by glucose on reaction with CH3OH and dry HCl can be explained in terms of the cyclic structure. The equilibrium mixture of a-and (3-glucose react separately with methanol in the presence of dry HCl gas to form the corresponding methyl D-glucosides.

These pentaacetates donot have a free -OH group at C1 and hence are not hydrolysed in aqueous solution to produce the open chain aldehyde form and hence do not react with NH2OH to form glucose oxime.

(v) The existence of glucose in two crystalline forms termed as a and β-D-glucose can again be explained on the basis of cyclic structure of glucose and not by its open chain structure. It was proposed that one of the -OH groups may add to – CHO group and form a cyclic hemiacetal structure. It was found that glucose forms a 6-membered ring in which -OH at C – 5 is involved in ring formation. This explains the absence of -CHO group and also existence of glucose in two forms as shown below. These two forms exist in equilibrium with open chain structure.

Question 11.
What are essential and non-essential amino acids? Give two examples of each type.
There are about 20 amino acids which make up the bio-proteins. Out of these 10 amino acids (non-essential) are synthesised by our bodies and rest are essential in the diet (essential amino acids) and supplied to our bodies by food which we take because they cannot be synthesised in our body.
e.g. Essential amino acid – Valine and Leucine
Non-essential amino acid – Glycine and Alanine

Question 12.
Define the following as related to proteins

(i) Peptide Linkage : Proteins are the polymers of a-amino acids which are connected to each other by peptide bond or peptide linkage. Chemically, peptide linkage is an amide formed between -COOH group and -NH2 group. The reaction between two molecules of similar or different amino acids, proceeds through the combination of the amino group of one molecule with the carboxyl group of the other. This results in the elimination of a water molecule and formation of a peptide bond -CO-NH-. The product of the reaction is called a dipeptide because it is made up of two amino acids. For example, when carboxyl group of glycine combines with the amino group of alanine we get a dipeptide, glycylalanine.

(ii) Primary Structure : Proteins may have one or more polypeptide chains. Each polypeptide in a protein has amino acids linked with each other in a specific sequence and it is this sequence of amino acids that is said to be the primary structure of that protein. Any change in this primary structure i.e., the sequence of amine acids creates a different protein.

(iii) Denaturation : Protein found in a biological system with a unique three-dimensional structure and biological activity is called a native protein. When a protein in its native form, is subjected to physical change like change in temperature or chemical change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled and protein loses its biological activity. This is called denaturation of protein. During denaturation 2° and 3° structures are destroyed but 1° structure remains intact. The coagulation of egg white on boiling is a common example of denaturation. Another example is curdling of milk which is caused due to the formation of lactic acid by the bacteria present in milk.

Question 13.
What are the common types of secondary structure of proteins?
The secondary structure of protein refers to the shape in which a long polypeptide chain can exist. They are found to exist in two different types of structures viz, a-helix and P-pleated sheet structure. These structures arise due to the regular folding of the backbone of the polypeptide chain due to hydrogen bonding between

Question 14.
What type of bonding helps in stabilising the α-helix structure of proteins?
α-Helix is one of the most common ways in which a polypeptide chain forms all possible hydrogen bonds by twisting into a right handed screw (helix) with the -NH group of each amino acid residue hydrogen bonded to the >C = O of an adjacent turn of the helix.

Question 15.
Differentiate between globular and fibrous proteins.
Characteristic differences between globular and fibrous proteins can be given as :

Globular Proteins

  1. These are cross linked proteins and are condensation product of acidic and basic amino acids.
  2. These are soluble in water, mineral acids and bases.
  3. These proteins have three dimensional folded structure. These are stabilised by internal hydrogen bonding, e.g., egg albumin enzymes.

Fibrous Proteins

  1. These are linear condensation polymer
  2. These are insoluble in water but soluble in strong acids and bases.
  3. These are linear polymers held together by intermolecular hydrogen bonds. e.g., hair, silk.

Question 16.
How do you explain the amphoteric behaviour of amino acids?
Due to dipolar or Zwitter ionic structure, amino acids are amphoteric in nature. The acidic character of the amino acids is due to the N + H3 group while the basic character is due to the COO – group.

Question 17.
What are enzymes?
Life is possible due to the coordination of various chemical processes in living organisms. An example is the digestion of food, absorption of appropriate molecules and ultimately production of energy. This process involves a sequence of reactions and all these reactions occur in the body under very mild conditions. This occurs with the help of certain biocatalysts called enzymes. Almost all the enzymes are globular proteins. Enzymes are specific for a particular reaction and for a particular substrate. They are generally named after the compound or class of compounds upon which they work. For example, the enzyme that catalyses hydrolysis of maltose into glucose is named as maltose.

Question 18.
What is the effect of denaturation on the structure of proteins?
Proteins are very sensitive to the action of heat, mineral acids, alkalies etc. On heating or on treatment with mineral acids, soluble forms of proteins such as globular proteins often undergo coagulation or precipitation to give fibrous proteins which are insoluble in water. This coagulation also results in the loss of the biological activity of the protein. That is why the coagulated proteins so formed are called denatured proteins.. Chemically, denaturation does not change the primary structure but brings about changes in the secondary and tertiary structure of proteins.

Question 19.
How are vitamins classified? Name the vitamin responsible for the coagulation of blood.
Vitamins are classified into two groups depending upon their solubility in water or fat.
(i) Fat soluble vitamins : Vitamins which are soluble in fats and oils but insoluble in water are kept in this group. These are vitamins A, D, E and K. They are stored in liver and adipose (fat storing) tissues.
(ii) Water soluble vitamins : B group vitamins and vitamin C are soluble in water so they are grouped together. Water soluble vitamins must be supplied regularly in diet
because they are readily excreted in urine and cannot be stored (except vitamin B12) in our body.
Vitamin K is responsible for coagulation of blood.

Question 20.
Why are vitamin A and vitamin C essential to us? Give their important sources.
Deficiency of vitamin A causes Xerophthalmia (hardening of cornea of the eye) and night blindness. So its use is essential to us. It is available in fish liver oil, carrots, butter and milk. It promotes growth and increases resistance to diseases. Vitamin C is very essential to us because its deficiency causes Scurvy (bleeding of gums) and pyorrhea (loosening and bleeding of teeth). Vitamin C increases resistance of the body towards diseases. Maintains healthy skin and helps cuts and abrasions to heat properly. It is soluble in water. It is present in citrus fruits, e.g.,oranges, lemons, amla, tomato. green vegetables (Cabbage) chillies, sprou pulses and germinated grains.

Question 21.
What are nucleic acids? Mention their two important functions.
Nucleic acids : They constitute an important class of biomolecules which are found in the nuclei of all living cells in the form of nucleoproteins (i.e., proteins containing nucleic acid as the prosthetic group). Nucleic acids are the genetic materials of the cells and are responsible for transmission of hereditary effect from one generation to the other and also carry out the biosynthesis of proteins. Nucleic acids are biopolymers (i.e., polymers present in the living system). The genetic information coded in nucleic acids controls the structure of all proteins including enzymes and thus governs the entire metabolic activity in the living organism.
Two important functions of nucleic acids are :

  1. Replication : The process by which a single DNA molecule produces two identical copies of itself is called replication.
  2. Protein Synthesis : DNA may be regarded as the instrument manual for the synthesis of all proteins present in the cell.

Question 22.
What is the difference between a nucleoside and a nucleotide?
Nucleoside : A nucleoside contains only two basic components of nucleic acids, i.e., a pentose sugar and a nitrogenous base. It may be represented as Sugar-base. Depending upon the type of sugar present, nucleosides are of two types :

Nucleotides : A nucleotide contains all the three basic components of nucleic acids, i.e., a phosphoric acid group, a pentose sugar and a nitrogenous base. In other words, nucleotides are nucleoside monophosphates.

Depending upon the type of sugar present, nucleotides like nucleosides are of two types :

Question 23.
The two strands in DNA are not identical but are complementary. Explain.
Two nucleic acid chains are wound about each other and held together by hydrogen bonds between pairs of bases. The trands are complementary to each other because the hydrogen bonds are formed between specific pairs of bases. Adenine forms hydrogen bonds with thymine whereas cytosine forms hydrogen bonds with guanine.

Question 24.
Write the important structural and functional differences between DNA and RNA.
Difference between DNA and RNA.

Question 25.
What are the different types of RNA found in the cell?
RNA molecules are of three types and they perform different functions. They are named as messenger RNA (m-RNA), ribosomal RNA (r-RNA) and transfer RNA (t-RNA).
Now that you are provided all the necessary information regarding NCERT Solutions for Class 12 Chemistry Chapter 14 Biomolecules and we hope this detailed NCERT Solutions are helpful.

Part 3: The Central Dogma

To understand why DNA is so crucial to our cells, we need to understand the central dogma of biology. Although the central dogma might sound daunting, it simply describes the relationship between DNA, RNA, and protein. The central dogma of biology states that DNA creates RNA, which in turn creates protein. We’ve drawn this relationship below for you.

The central dogma explains how genetic information travels from storage in DNA to an intermediary RNA and finally is translated into a protein that can cause physical and chemical changes in the cell.

There are two necessary steps our cells must take to use the genetic information stored in their DNA. These steps are called transcription and translation. The cell works hard to pass on the encoded genetic information accurately at each step.

A) Transcription

Transcription is the first step of the central dogma. Transcription describes the process by which an RNA transcript is created from existing DNA. DNA and RNA use similar, but slightly different languages to encode genetic information. DNA encodes its information using four nucleotides: adenine, guanine, cytosine, and thymine. RNA utilizes the same nucleotides, with the exception of uracil in place of thymine. For more information, be sure to refer to our guide on RNA.

In transcription, an enzyme known as RNA polymerase acts as a translator to create a new RNA transcript. Single-copy DNA, or DNA regions that encode proteins, are transcribed by RNA polymerase. (This is in contrast to repetitive DNA, which are long regions of repetitive sequences that act as introns or protective sequences and do not encode any proteins.) The transcription of a DNA region may also be affected by transcription factors, proteins that bind to a segment of DNA to either promote or repress its transcription.

Similar to DNA polymerase, the RNA polymerase can read DNA bases and translate them into the complementary RNA sequence. Since DNA is found in the cell nucleus, that is where RNA polymerase does its work.

RNA polymerase finishes translating when it encounters a stop sequence. If all goes well, the cell has a newly synthesized strand of RNA, complementary to the template strand of the DNA and identical to the coding strand. Later on, the RNA can be used to create proteins.

B) Translation

Translation is the second step of the central dogma. Translation describes how proteins are created from synthesized RNA transcripts. Before translation can occur, synthesized RNA transcripts undergo posttranscriptional processing and are sent out of the cell nucleus. You can find more about posttranscriptional processing in our study guide on RNA.

Once the cell has mRNA transcripts, translation starts. Translation occurs in the ribosomes of cells. The ribosome binds to the mRNA transcripts and reads the codons in the RNA sequence. A codon is simply a sequence of three nucleotide bases. Each codon specifies a specific amino acid in the protein sequence. For example, the nucleotide sequence adenine, uracil, guanine represents the codon, AUG. If we look at a codon table, we’ll see that AUG specifies the amino acid methionine. The AUG codon is a start or initiation codon, which signals the ribosome to begin translation. There are additional stop codons (UAA, UGA, UAG) that signal the termination of translation.

You may note that there are more combinations of codons that are possible than there are amino acids. While there are 43, or 64 codons that can be formed by nucleotides, there are only 21 translated amino acids. As a result, multiple codons may code for the same amino acid. This is a phenomenon known as degeneracy.

For example, the amino acid alanine can be encoded by multiple codons: including GCU, GCA, GCC, GCG. You may note that while these four codons all share identical nucleotides in the first two positions, the third nucleotide does not seem to be as important. This is made possible by the wobble effect, or non-Watson Crick base pairing that allows for weak binding between the third nucleotide of the codon in mRNA and anticodon in tRNA. Thus, multiple codons can translate to the same amino acid—without requiring the presence of 64 unique tRNA molecules!

Each of the codons with an mRNA sequence has a complementary anticodon. tRNA molecules contain anticodon sequences that are able to bind their complementary codons, thus facilitating the interaction between tRNA and the ribosome. For more information on this process, be sure to refer to our guide on RNA.

Errors can occur during DNA replication, transcription, or translation. When mutations occur, the translated meaning of a codon can also be changed. A missense mutation results in the translation of a different amino acid. A nonsense mutation results in the translation of a premature stop codon, effectively truncating the translated protein.

Once the ribosome finishes reading the sequence, it has assembled a polypeptide of amino acids that can then fold into a protein.

Proteins are the expression of the genetic information contained in the DNA. Proteins carry out different functions throughout the cell based on their structure. Some proteins, such as cell membrane receptors, attach to the cell membrane and relay cell signals, while other proteins act as catalysts, breaking down waste products. It’s important to remember that while DNA might contain the information our cells need to function, proteins carry out the processes vital to the cell’s survival. You can find more information in our study guides on proteins and enzymatic function.

C) Exceptions to the central dogma

For the most part, the central dogma describes how genetic information is passed linearly from DNA to RNA to protein. There is, however, one large exception: retroviruses.

Retroviruses are RNA viruses that use a special enzyme known as reverse transcriptase to create a double-stranded DNA molecule out of the RNA they possess. Recall that the central dogma that RNA is produced using DNA as a template. Retroviruses violate this rule instead, DNA is produced using RNA.

Human immunodeficiency virus, or HIV, is one such example of a retrovirus. Once HIV infects a cell, the virus uses a reverse transcriptase to create a DNA molecule from its RNA. The virus then uses machinery within the host cell’s nucleus to splice newly synthesized viral DNA into the host cell’s genome. This means that the HIV virus can be efficiently transcribed using the cell’s own machinery. HIV also cannot be cured without genomic editing techniques, as viral proteins would be transcribed anytime host proteins are produced.

We’ve updated the central dogma diagram from above to illustrate how retroviruses go against the linear flow of genetic information.

Watch the video: The definition to 5 end and 3 end of a DNA strand - Simple animated HD (July 2022).


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