Biomolecules (Biology)

NEET Botany notes. RelaxPrep is not affiliated with the NTA or the IITs.

Revision sheet

Cell chemistry is studied by grinding tissue in cold trichloroacetic acid and straining it: the filtrate is the acid-soluble pool (small molecules), the residue is the acid-insoluble fraction (macromolecules, plus lipids that arrive as membrane fragments).
Burning tissue leaves ash, which is the inorganic part; carbon, hydrogen, oxygen and nitrogen leave as gases. Water is the most abundant compound in a cell; protein is the most abundant organic compound.
Every standard amino acid has an amino group, a carboxyl group, a hydrogen and a variable R group on the same ($\alpha$) carbon. In a zwitterion the groups are $\ce{NH3+}$ and $\ce{COO-}$.
A peptide bond is an amide link formed by condensation between the carboxyl of one amino acid and the amino group of the next, releasing one water. A chain of $n$ residues has $n-1$ peptide bonds.
Protein levels: primary (sequence), secondary ($\alpha$-helix, $\beta$-sheet, held by backbone H bonds), tertiary (3-D fold of one chain, held by R-group interactions), quaternary (several chains together, as in haemoglobin).
Denaturation destroys secondary, tertiary and quaternary structure but leaves the peptide bonds, and so the primary structure, intact.
Starch and glycogen are $\alpha$-glucose polymers; cellulose is a $\beta$-1,4 glucose polymer; inulin is a fructose polymer; chitin is built from an amino sugar (N-acetylglucosamine). Sucrose is a non-reducing disaccharide.
Fats and oils are triglycerides (glycerol plus three fatty acids, joined by ester bonds). Phospholipids are amphipathic and form the membrane bilayer. Lipids are not true macromolecules.
Nucleoside = base + sugar; nucleotide = base + sugar + phosphate. Purines are A and G; pyrimidines are C, T and U. DNA has T and deoxyribose; RNA has U and ribose.
In double-stranded DNA the strands are antiparallel; A pairs with T by two H bonds, G with C by three. $A+G = T+C$ always.
Enzymes are mostly proteins (some are RNA). They lower the activation energy; they do not change $\Delta G$ or the equilibrium position, and they are not used up.
$K_m$ is the substrate concentration at which $V = V_{max}/2$. A low $K_m$ means high apparent affinity. $V_{max}$ is proportional to the amount of enzyme; $K_m$ is not.
Competitive inhibitor: $V_{max}$ unchanged, apparent $K_m$ rises, relieved by more substrate. Non-competitive: $V_{max}$ falls, $K_m$ unchanged.
Six enzyme classes in order: oxidoreductases, transferases, hydrolases, lyases, isomerases, ligases.
Apoenzyme (protein) + cofactor = holoenzyme. Cofactors are prosthetic groups (tightly bound, such as haem), coenzymes (organic, often vitamin-derived, such as NAD) and metal ions.
Metabolism = anabolic (building, uses energy) + catabolic (breaking down, releases energy) reactions. A living cell is a steady-state open system, not an equilibrium system.
Primary metabolites have clear roles in growth and normal physiology; secondary metabolites (alkaloids, terpenoids, pigments, rubber, toxins and so on) have ecological roles and many human uses.

Michaelis-Menten equation: $V = \dfrac{V_{max}[S]}{K_m + [S]}$ ($V$ in concentration per unit time; $K_m$ in the same units as $[S]$.)
Substrate needed for a fraction of $V_{max}$: $[S] = K_m\,\dfrac{f}{1-f}$ ($f = V/V_{max}$; for example $f = 0.9$ needs $[S] = 9K_m$.)
Lineweaver-Burk form: $\dfrac{1}{V} = \dfrac{K_m}{V_{max}}\cdot\dfrac{1}{[S]} + \dfrac{1}{V_{max}}$ (Straight line: y-intercept $1/V_{max}$, x-intercept $-1/K_m$, slope $K_m/V_{max}$.)
Competitive inhibition, apparent $K_m$: $K_m^{app} = K_m\left(1 + \dfrac{[I]}{K_i}\right)$ ($V_{max}$ is unchanged.)
Maximum velocity: $V_{max} = k_{cat}[E]_{total}$ (Doubling the enzyme doubles $V_{max}$ and leaves $K_m$ unchanged.)
Mass of a linear peptide: $M_{peptide} = \sum M_{amino\ acids} - 18\,(n-1)$ ($n$ residues; 18 is the molar mass of water in g/mol.)
Hydrogen bonds in double-stranded DNA: $H = 2\,N_{A=T} + 3\,N_{G\equiv C}$ (Counted in base pairs.)
Isoelectric point, neutral amino acid: $pI = \dfrac{pK_{a1} + pK_{a2}}{2}$ (For acidic or basic side chains use the two $pK_a$ values on either side of the neutral form.)

Topics

Chemical analysis of living tissue

There are two kinds of analysis, and they answer different questions. To learn which elements a tissue contains, a weighed piece is dried and then burned: carbon, hydrogen, oxygen and nitrogen escape as gases, and the residue, the ash, holds the mineral elements such as calcium, magnesium, sodium, potassium and phosphorus. To learn which compounds it contains, the tissue is instead crushed in cold trichloroacetic acid and the resulting paste is passed through a cloth filter.

What runs through the cloth is the acid-soluble pool: small molecules of roughly 18 to 800 Da, such as amino acids, simple sugars, nucleotides, organic acids and ions. What stays on the cloth is the acid-insoluble fraction, which holds the macromolecules (proteins, polysaccharides and nucleic acids, usually 10,000 Da or more). Lipids weigh under about 800 Da and so are not macromolecules, yet they are found on the cloth too: crushing tears the cell membranes into vesicles that do not dissolve in the aqueous acid, and the lipids travel with them. These size limits are teaching conventions, not sharp physical boundaries.

In terms of mass, water is the biggest component of a cell, at about 70 to 90 per cent. Protein is the largest organic share (about 10 to 15 per cent), and nucleic acids, carbohydrates, lipids and ions make up the rest, with ions only about 1 per cent. Plants, animals and microbes yield the same classes of compound, which gives biology a shared chemical vocabulary.

Key points

Elemental analysis: dry, then burn; the ash is the inorganic fraction.
Filtrate = acid-soluble pool = micromolecules (about 18 to 800 Da). Retentate = acid-insoluble fraction = macromolecules plus membrane-derived lipids.
Biomacromolecules in the retentate: proteins, nucleic acids and polysaccharides. They are polymers; lipids are not.
Water is the most abundant compound in a cell; protein is the most abundant organic compound.
Micromolecules in the filtrate are the building blocks and intermediates (amino acids, sugars, nucleotides) that the macromolecules are made from.

Common traps

Lipids are found in the acid-insoluble fraction although they are small molecules. Their position there is a side effect of membrane fragmentation, not a sign that they are polymers.
The ash of a burnt tissue is not the whole of the inorganic content of the living cell: water and dissolved carbon dioxide were part of the living tissue too, and they leave during drying and burning.
The acid-soluble pool is not 'the cytoplasm' and the acid-insoluble fraction is not 'the membranes': the split is made by size and solubility, and each pool comes from every part of the cell.

Amino acids and zwitterions

Amino acids are organic compounds in which one carbon, the $\alpha$ carbon, carries four different substituents: an amino group, a carboxyl group, a hydrogen atom and a side chain called R. Proteins are built from twenty standard amino acids, and R is the only thing that differs between them. Glycine has R = H, alanine has a methyl group ($\ce{CH3}$) and serine has a hydroxymethyl group ($\ce{CH2OH}$). Because the $\alpha$ carbon of every amino acid except glycine has four different groups, it is a chiral centre, and proteins use only the L-form.

The side chain decides the character of the amino acid. Aspartic and glutamic acids carry an extra carboxyl group (acidic); lysine, arginine and histidine carry extra basic groups; valine, leucine and alanine are neutral and non-polar; phenylalanine, tyrosine and tryptophan are aromatic; cysteine carries a thiol that can pair up in a disulfide bridge; proline is unusual because its side chain loops back to the nitrogen, making it a cyclic imino acid. Animals cannot make about half of the standard set and must take them in food; these are the essential amino acids.

In water an amino acid is not the neutral molecule drawn in many books. The carboxyl group gives up a proton and the amino group accepts one, so the molecule carries $\ce{NH3+}$ and $\ce{COO-}$ at once; this dipolar ion is a zwitterion. It can act as an acid or as a base, so amino acids are amphoteric. In strong acid the molecule carries a net positive charge, in strong base a net negative charge, and at one intermediate pH, the isoelectric point (pI), the net charge is zero and the amino acid migrates neither way in an electric field.

Key points

General structure: $\alpha$ carbon bonded to $\ce{NH2}$, $\ce{COOH}$, H and R.
Standard proteins use 20 amino acids; they differ only in R.
Glycine (R = H) is the only achiral standard amino acid; protein amino acids are L-isomers.
Acidic: aspartic acid, glutamic acid. Basic: lysine, arginine, histidine. Aromatic: phenylalanine, tyrosine, tryptophan. Sulfur-containing: cysteine, methionine.
Amino acids are amphoteric and exist as zwitterions near neutral pH; they have a pI at which net charge is zero.
Essential amino acids must come from the diet; non-essential ones can be made by the body.

Isoelectric point of a neutral amino acid: $pI = \dfrac{pK_{a1} + pK_{a2}}{2}$ (Average of the carboxyl and amino $pK_a$ values; glycine: $(2.34 + 9.60)/2 \approx 5.97$.)

Common traps

'Essential' describes diet, not importance: every standard amino acid is needed for protein synthesis; an essential amino acid is one the animal cannot synthesise.
Proline is called an imino acid because its nitrogen is bonded to two carbons, but it still forms a normal peptide bond and is one of the twenty protein amino acids.
Amino acids are not neutral molecules in solution: the uncharged $\ce{NH2}$/$\ce{COOH}$ form is only a drawing convention; the zwitterion is the dominant form near neutral pH.

Shortcuts

Finding the pI of acidic and basic amino acids. Average the two $pK_a$ values that bracket the neutral species. Neutral amino acid: $pI = (pK_{a1}+pK_{a2})/2$. Aspartic acid: the two carboxyl values $1.88$ and $3.65$ give $pI \approx 2.77$. Lysine: the two amino values $8.95$ and $10.53$ give $pI \approx 9.74$. Acidic residues therefore have a low pI and basic ones a high pI.

Peptide bond and primary structure

When the carboxyl group of one amino acid meets the amino group of another, a molecule of water is removed and a C-N bond forms between the carbonyl carbon and the nitrogen. This amide link is the peptide bond, and the reaction is a condensation (dehydration synthesis). In cells it is catalysed on the ribosome and driven by the energy of ATP and GTP; the reverse reaction, hydrolysis, is carried out by peptidases and proteases.

A short chain is a peptide (di-, tri-, oligo-), and a long chain is a polypeptide; a protein is one or more polypeptides folded into working shape. A linear chain has direction: one end keeps a free amino group (the N-terminus), the other a free carboxyl group (the C-terminus), and sequences are always written from N to C. The peptide bond has partial double-bond character because the nitrogen lone pair is shared with the carbonyl group, so the six atoms around it lie in one plane and rotation about the C-N bond is restricted. The freedom of a chain therefore lies in rotations about the bonds at the $\alpha$ carbons.

The order of amino acids is the primary structure, and it is fixed by the gene. It matters more than any single residue: replacing one amino acid with another can change folding and function. In sickle-cell haemoglobin one glutamic acid in the $\beta$ chain is replaced by valine, and the result is a protein that polymerises when deoxygenated. With twenty choices at every position, a chain of length $n$ can have $20^n$ different sequences, which is why so many distinct proteins are possible.

Key points

Peptide bond = amide link, formed by loss of one $\ce{H2O}$ (condensation); broken by hydrolysis.
A linear peptide of $n$ amino acids has $n-1$ peptide bonds and lost $n-1$ water molecules.
Chains are read N-terminus to C-terminus; a linear chain has one free $\ce{NH2}$ end and one free $\ce{COOH}$ end.
Partial double-bond character makes the peptide group planar and rigid.
Primary structure is the amino acid sequence only; peptide bonds are the only covalent bonds in the backbone.

Mass of a linear peptide: $M = \sum M_{aa} - 18\,(n-1)$

Common traps

A circular peptide has as many peptide bonds as residues ($n$, not $n-1$) and no free terminus.
Peptide bonds are covalent and survive denaturation; boiling an egg does not break them. Digestive enzymes or strong acid are needed to hydrolyse them.
The side-chain amino or carboxyl groups of lysine or glutamic acid are not part of the backbone peptide bonds, so they stay free in the chain.

Shortcuts

Counting bonds, water and sequences. For a linear peptide of $n$ residues: peptide bonds $= n-1$, water lost $= n-1$, mass $= \sum M_{aa} - 18(n-1)$, number of possible sequences from 20 amino acids $= 20^n$ (so $20^3 = 8000$, $20^4 = 160000$).

Levels of protein structure, functions and denaturation

The primary structure folds into repeating local patterns that make up the secondary structure. In an $\alpha$-helix the chain coils clockwise (right-handed) with about 3.6 residues per turn; each C=O group forms a hydrogen bond with the N-H group four residues further along, so the H bonds run roughly parallel to the helix axis. In a $\beta$-pleated sheet, stretches of chain lie side by side, either parallel or antiparallel, and the H bonds link neighbouring strands. All these H bonds involve the backbone, not the R groups.

Tertiary structure is the full three-dimensional shape of one polypeptide. It forms as the secondary elements pack together, and it is held by R-group interactions: hydrophobic side chains cluster in the interior, away from water; hydrogen bonds, ionic (salt) bridges and van der Waals contacts hold the fold; and covalent disulfide bonds between cysteine residues lock parts of it together. The active site of an enzyme and the haem pocket of globin are features of the tertiary structure. Quaternary structure exists only in proteins that contain more than one polypeptide: adult haemoglobin has four chains, two $\alpha$ and two $\beta$, each carrying a haem group, held together by the same kinds of weak interaction.

Because function depends on shape, anything that disrupts the weak interactions can destroy activity. This is denaturation: heat, extreme pH, urea, organic solvents or heavy metals unfold the protein, so the secondary, tertiary and quaternary levels are lost, but the peptide bonds remain. Mild denaturation can be reversed if the sequence still contains the information to refold; boiled egg white shows an irreversible case. Proteins are the most diverse class of biomolecule in function: collagen is the most abundant protein in the animal world and RuBisCO is the most abundant protein in the whole biosphere; enzymes such as trypsin, hormones such as insulin, antibodies, receptors and transporters (GLUT-4) are all proteins.

Key points

Primary: sequence. Secondary: $\alpha$-helix and $\beta$-sheet (backbone H bonds). Tertiary: 3-D shape of one chain. Quaternary: arrangement of two or more chains.
$\alpha$-helix: right-handed, about 3.6 residues per turn, pitch about 0.54 nm; the rise per residue is about 0.15 nm.
Tertiary structure bonds: hydrophobic interactions, H bonds, ionic bonds, van der Waals forces, disulfide ($-$S$-$S$-$) bridges.
Haemoglobin: $\alpha_2\beta_2$ tetramer, four haem groups, an example of quaternary structure. Myoglobin is a single chain (no quaternary structure).
Denaturation breaks levels 2, 3 and 4 but not level 1.
Collagen is the most abundant animal protein; RuBisCO is the most abundant protein on Earth.

Rise per turn of an $\alpha$-helix: $3.6 \times 0.15\ \text{nm} \approx 0.54\ \text{nm}$ (Residues per turn times rise per residue.)

Common traps

Quaternary structure is not present in every protein; a protein made of a single chain stops at tertiary structure.
Disulfide bonds are covalent, but they are not peptide bonds and they are not part of the primary structure.
A secondary structure depends on backbone N-H and C=O groups, so an $\alpha$-helix can form from many different amino acid sequences.
Denaturation is a change in shape; it is not hydrolysis and it does not change the amino acid sequence.

Shortcuts

Which bond belongs to which level. Level 1 uses only peptide (covalent) bonds. Level 2 uses only backbone H bonds. Level 3 adds R-group interactions, including S-S bridges. Level 4 uses the same R-group interactions between chains. A question that says 'only H bonds between backbone atoms' is pointing to level 2.

Monosaccharides, disaccharides and the glycosidic bond

Carbohydrates are polyhydroxy aldehydes or ketones, or compounds that give them on hydrolysis. The monosaccharides are the units: glucose is an aldohexose, fructose is a ketohexose, ribose and deoxyribose are aldopentoses. In water they close into rings, and the new hydroxyl at the anomeric carbon (C-1 in glucose, C-2 in fructose) can point either way, giving the $\alpha$ and $\beta$ forms. That small difference decides whether the polymer is a storage compound or a structural one.

Two monosaccharides join by a glycosidic bond, formed by condensation between the anomeric hydroxyl of one and a hydroxyl of the other. Maltose is glucose + glucose joined $\alpha$-1,4. Lactose, the milk sugar, is galactose + glucose joined $\beta$-1,4. Sucrose is glucose + fructose joined through both anomeric carbons ($\alpha$-1 to $\beta$-2). A sugar is called reducing if it has a free anomeric carbon that can open to an aldehyde or a ketol and reduce Benedict's or Fehling's reagent; maltose and lactose are reducing, but sucrose is not, because both anomeric carbons are used in the link.

Boiling sucrose with dilute acid, or treating it with the enzyme invertase, splits it into equal amounts of glucose and fructose, which together are called invert sugar. The name comes from the optical rotation, which changes from dextro (positive) to laevo (negative) because fructose rotates more strongly to the left than glucose does to the right.

Key points

Monosaccharides: glucose and galactose (aldohexoses), fructose (ketohexose), ribose and deoxyribose (aldopentoses).
Disaccharides: sucrose (glucose + fructose), maltose (glucose + glucose), lactose (galactose + glucose).
Glycosidic bond: formed by condensation, broken by hydrolysis.
Sucrose is non-reducing; maltose and lactose are reducing. Benedict's and Fehling's tests detect reducing sugars.
Hydrolysis of sucrose gives invert sugar, a 1:1 mixture of glucose and fructose.

Common traps

Sucrose does not give a positive Benedict's test before hydrolysis, although both of its monosaccharides are reducing sugars when free.
Lactose is galactose + glucose, not glucose + glucose; the latter is maltose.
Glycosidic bonds join sugars; glycosidic bonds are not peptide or ester bonds even though all three come from condensation.

Polysaccharides: starch, glycogen, cellulose, chitin, inulin

A polysaccharide is a long chain of monosaccharides joined by glycosidic bonds. Each chain has two different ends: one end has a free anomeric carbon (the reducing end), the other does not (the non-reducing end). A polymer made of one kind of unit is a homopolymer. Starch, glycogen and cellulose are homopolymers of glucose; inulin is a homopolymer of fructose.

Starch is the storage form in plants. It has two parts: amylose, an unbranched chain of $\alpha$-1,4 linked glucose that coils into a helix, and amylopectin, which has the same chain with $\alpha$-1,6 branches every 20 to 30 units or so. Glycogen is the storage form in animals and fungi; it has the same type of bonds as amylopectin but is branched more often (every 8 to 12 units), so it is more compact and offers many chain ends for quick release of glucose. Iodine turns starch blue-black; cellulose gives no such colour.

Cellulose is built from $\beta$-1,4 linked glucose. Because every second unit is turned over, the chain is straight and ribbon-like, and chains pack side by side with H bonds into strong fibrils that make plant cell walls. Humans lack an enzyme that cuts $\beta$-1,4 bonds, so cellulose passes as fibre, whereas amylase cuts $\alpha$-1,4 bonds in starch. Chitin, the polymer in the cell walls of fungi and the exoskeleton of arthropods, is built from an amino-sugar derivative (N-acetylglucosamine) rather than glucose itself. Peptidoglycan in bacterial walls contains amino sugars linked to short peptides.

Key points

Starch: plants; amylose ($\alpha$-1,4, unbranched) + amylopectin ($\alpha$-1,4 and branches at $\alpha$-1,6).
Glycogen: animals and fungi; $\alpha$-1,4 with $\alpha$-1,6 branches every 8 to 12 residues (more branched than amylopectin).
Cellulose: plants; $\beta$-1,4 glucose; most abundant organic compound in the biosphere.
Inulin: fructose polymer. Chitin: amino-sugar polymer (fungal cell wall, arthropod exoskeleton).
Iodine test: starch gives blue colour; cellulose does not.

Common traps

Cellulose is built from glucose just as starch is; the difference is the $\beta$ configuration, not the monomer.
Chitin is not a homopolymer of glucose, so a question asking for the polysaccharide that is not made from glucose is pointing to chitin or inulin.
Branching points are $\alpha$-1,6 bonds; the main chain remains $\alpha$-1,4.
A polysaccharide molecule has only one reducing end, however long it is, so polysaccharides are not tested as reducing sugars.

Shortcuts

Alpha versus beta glucose polymers. Starch and glycogen are $\alpha$-glucose polymers that animals digest with amylase; cellulose and chitin are $\beta$-linked and are not digested by it. For a branched glucose polymer, the more frequent the branches (every 8 to 12 versus every 20 to 30 units), the closer it is to glycogen rather than amylopectin.

Lipids: fatty acids, glycerides, phospholipids, steroids

Lipids are a group defined by solubility, not by structure: they dissolve in organic solvents and not in water. Fatty acids are the simplest: a carboxyl group at the end of a hydrocarbon chain. Palmitic acid has 16 carbons with no double bonds; arachidonic acid has 20 carbons and four double bonds. A saturated chain is straight and packs tightly, whereas each cis double bond puts a bend in the chain and lowers the melting point.

Glycerol is a three-carbon alcohol with three hydroxyl groups. When one, two or three of them are esterified with fatty acids the product is a mono-, di- or triglyceride. Fats (solid at room temperature) and oils (liquid, like gingelly oil) are triglycerides; oils contain more unsaturated chains. Triglycerides are the long-term energy store: the energy yield is about 9 kcal per gram, against about 4 for carbohydrate or protein. A phospholipid replaces one fatty acid by a phosphate group joined to a nitrogenous group such as choline, as in lecithin. It has a polar head and two non-polar tails, so it is amphipathic, and in water it forms the bilayer of the cell membrane.

Steroids are lipids built on four fused carbon rings; cholesterol is a component of animal cell membranes and the starting material for steroid hormones. Because lipids are small molecules (below about 800 Da) and not repeating polymers, they are not counted among biomacromolecules, yet they appear in the acid-insoluble fraction of the tissue analysis because of membrane fragments.

Key points

Fatty acid = $\ce{R-COOH}$; R has 1 to 19 carbons. Saturated: no double bond. Unsaturated: one or more double bonds.
Glycerol + 3 fatty acids $\rightarrow$ triglyceride + 3 $\ce{H2O}$ (three ester bonds).
Phospholipid: glycerol + 2 fatty acids + phosphate + a nitrogenous group; amphipathic.
Fats are solid at room temperature; oils are liquid because of unsaturation.
Cholesterol is a steroid; it is part of animal cell membranes.
Lipids are not macromolecules: they have low molecular mass and are not polymers.

Common traps

Lipids are found in the acid-insoluble fraction of a tissue analysis only because of membrane fragmentation, not because they are macromolecules.
The ester link in a glyceride is not the same as the glycosidic or peptide link; the ester forms between an acid ($\ce{-COOH}$) and an alcohol ($\ce{-OH}$).
Phospholipids are amphipathic: they have both polar and non-polar ends, which makes them different from a triglyceride, which is wholly non-polar.

Shortcuts

Water count in condensation. Each bond formed by condensation releases one $\ce{H2O}$. A triglyceride from glycerol and three fatty acids therefore releases 3 water molecules, a dipeptide releases 1, and a linear peptide of $n$ residues releases $n-1$. Energy density: fat about 9 kcal/g, carbohydrate and protein about 4 kcal/g.

Nucleosides, nucleotides and nitrogenous bases

A nucleotide has three parts: a nitrogenous base, a pentose sugar and a phosphate group. Strip off the phosphate and what remains is a nucleoside. Nucleosides are written with the sugar and base combined: adenosine, guanosine, cytidine, uridine and thymidine; the matching nucleotides are the corresponding monophosphates (adenylic acid and so on).

The bases fall into two families. Purines (adenine and guanine) have a double ring, and pyrimidines (cytosine, thymine and uracil) have a single ring. The sugar is ribose in RNA and 2-deoxyribose in DNA. Thymine is found in DNA and uracil in RNA; cytosine, adenine and guanine are present in both. The base joins carbon 1 of the sugar through an N-glycosidic bond; the phosphate is esterified to carbon 5 (a phosphoester bond).

Nucleotides are more than polymer units. ATP is an adenine nucleotide with three phosphates and is the universal energy currency of the cell. NAD, NADP, FAD and coenzyme A each contain a nucleotide portion. Chains of nucleotides form nucleic acids, which carry hereditary information and direct protein synthesis.

Key points

Nucleoside = base + sugar; nucleotide = base + sugar + phosphate.
Purines: adenine, guanine. Pyrimidines: cytosine, thymine, uracil.
DNA: deoxyribose and A, G, C, T. RNA: ribose and A, G, C, U.
Base-sugar link: N-glycosidic bond at C-1. Sugar-phosphate link: phosphoester bond at C-5.
ATP, NAD and FAD contain nucleotide units.

Common traps

A nucleoside has no phosphate. A statement that gives a nucleoside a phosphate group is describing a nucleotide.
Thymine is 5-methyluracil; it is the pyrimidine of DNA, while uracil is the pyrimidine of RNA.
The base is attached to the sugar by an N-glycosidic bond, not by a phosphoester bond; the phosphoester bond joins the phosphate to the sugar.

Structure of DNA and RNA

In a nucleic acid chain, nucleotides are joined by phosphodiester bonds: the 3' hydroxyl of one sugar is linked through a phosphate to the 5' carbon of the next. The chain therefore has a 5' end and a 3' end, and sequences are read 5' to 3'. The phosphate groups give nucleic acids their acidic nature and negative charge.

DNA is a double helix of two antiparallel chains: one runs 5' to 3' while the other runs 3' to 5'. The bases are inside, the sugar-phosphate backbones outside. A pairs with T by two hydrogen bonds, and G pairs with C by three. The pairing rule means a purine always faces a pyrimidine, so the double helix has a constant width of about 2 nm. In B-DNA there are 10 base pairs per turn, a pitch of 3.4 nm and a rise of 0.34 nm per base pair. Chargaff's observations follow from the pairing: in double-stranded DNA A = T and G = C, so purines equal pyrimidines, but the ratio (A + T)/(G + C) differs between species.

RNA is usually single-stranded, uses ribose and uracil, and is made as mRNA (carries the message), tRNA (brings amino acids) and rRNA (part of ribosomes, and itself a ribozyme at the peptidyl transferase site). Single strands may fold back on themselves to form double-stranded stems, as in tRNA. The 2' hydroxyl on ribose makes RNA less stable than DNA, which has only hydrogen at 2'.

Key points

Phosphodiester bond: links 3' of one nucleotide to 5' of the next; chains are antiparallel in the double helix.
A=T (two H bonds), G$\equiv$C (three H bonds); A + G = T + C in double-stranded DNA.
B-DNA: 10 bp per turn, pitch 3.4 nm, rise 0.34 nm per bp, diameter about 2 nm.
RNA: single-stranded, ribose, U instead of T; mRNA, tRNA, rRNA.
Higher G + C content means more H bonds and a more heat-stable double helix.

H bonds in double-stranded DNA: $H = 2\,N_{A=T} + 3\,N_{G\equiv C}$ (Count base pairs, not single bases.)

Common traps

Chargaff's rules apply to double-stranded DNA. A single-stranded DNA or RNA molecule need not have A = T or A = U.
A and T (or U) are paired by two hydrogen bonds, not three; the pair with three bonds is G-C.
The two strands run in opposite directions; the same direction would be called parallel and does not occur in the standard double helix.
Phosphodiester bonds hold the backbone together; hydrogen bonds hold the strands together.

Shortcuts

Base percentages from a single value. If A is $x$ per cent of a double-stranded DNA, then T is also $x$, and G = C = $(100 - 2x)/2 = 50 - x$ per cent. Check by $A + G = 50\%$. Hydrogen bonds: $H = 2N_{A=T} + 3N_{G\equiv C}$ counted over base pairs.

Enzymes: nature and mechanism of action

Enzymes are biological catalysts. Almost all are proteins; a few, such as the RNA of the ribosome that forms peptide bonds, are catalytic RNAs. Like any catalyst, an enzyme speeds up a reaction without being used up, and it does not change the free energy difference between substrate and product, so the equilibrium position is the same with or without it. What changes is the speed: the enzyme lowers the activation energy, the height of the barrier between the substrate and the high-energy transition state. A fast enzyme such as carbonic anhydrase handles several hundred thousand to about a million substrate molecules per second.

The substrate binds to a small pocket or cleft in the enzyme's tertiary structure called the active site. The shape and chemistry of the site give enzymes their specificity. In the old lock-and-key picture the substrate fits like a key; in the induced-fit picture the site changes shape on binding, so the enzyme grips the transition state. The enzyme and substrate form an enzyme-substrate (ES) complex, bonds are made or broken, an enzyme-product complex forms, and the products leave, freeing the enzyme for the next molecule. Binding strains the substrate, holds reacting groups close and correctly oriented, and may supply acid-base or covalent help.

Because the enzyme is a protein, its activity depends on its shape. Anything that alters the shape, such as high temperature or extreme pH, alters the activity, which leads to the temperature and pH profiles described in the next topic. Enzyme names usually end in -ase and indicate the substrate (lipase) or the reaction (dehydrogenase).

Key points

Enzymes: mostly proteins, a few RNA (ribozymes). They are not consumed.
They lower activation energy; $\Delta G$ of the reaction and the equilibrium are unchanged.
Active site: a pocket in the 3-D structure that binds the substrate; it explains specificity.
Sequence: E + S $\rightleftharpoons$ ES $\rightarrow$ EP $\rightarrow$ E + P.
Models: lock-and-key and induced fit.

Common traps

An enzyme does not supply energy to a reaction; it provides a lower-energy path to the same products.
A catalyst does not shift the equilibrium. If a reaction is endergonic, an enzyme alone cannot make it spontaneous.
The activation energy is the barrier from the substrate to the transition state, not the overall energy difference between substrate and product.

Factors affecting enzyme activity and Michaelis-Menten kinetics

Temperature: within the range where the enzyme is stable, the rate rises with temperature, roughly doubling for each 10 degrees C rise; beyond the optimum the enzyme starts to denature and the rate falls steeply. Low temperature slows the enzyme but leaves its structure intact, so it is reversibly inactive, whereas high temperature that denatures the enzyme usually does so irreversibly. Taq polymerase from the thermophile Thermus aquaticus is stable at the high temperatures of PCR. pH: each enzyme has an optimum, such as about 1.5 to 2 for pepsin in the stomach and about 8 for trypsin in the intestine; away from the optimum the ionisation of active-site groups and the protein's shape change.

Substrate concentration: at fixed enzyme concentration, the velocity of the reaction rises with substrate, steeply at first and then more slowly, and approaches a ceiling called $V_{max}$ when almost all the enzyme molecules are occupied. The simplest equation for this is the Michaelis-Menten equation, $V = V_{max}[S]/(K_m + [S])$. $K_m$ is the substrate concentration at which the velocity is half of $V_{max}$. A small $K_m$ means that half-saturation is reached at low substrate concentration, which is usually read as high affinity of the enzyme for its substrate.

$V_{max}$ depends on the amount of enzyme ($V_{max} = k_{cat}[E]_{total}$), so doubling the enzyme doubles $V_{max}$ but leaves $K_m$ unchanged. $K_m$ is a property of an enzyme with a given substrate under given conditions of pH and temperature. $V_{max}$ is approached only asymptotically; no finite substrate concentration reaches it exactly. Taking reciprocals of the equation gives the Lineweaver-Burk (double reciprocal) line, which converts the curve into a straight line and is the easy way to read $V_{max}$ and $K_m$ from data.

Key points

Temperature: rate rises up to the optimum, falls beyond it because of denaturation; cold only slows and does not denature.
pH: bell-shaped activity curve with an optimum; pepsin about 1.5 to 2, salivary amylase about 6.8, trypsin about 8.
Substrate: hyperbolic curve to $V_{max}$; $K_m$ = $[S]$ at $V_{max}/2$.
Low $K_m$ = high affinity; $V_{max}$ is proportional to enzyme concentration; $K_m$ is independent of it.
Lineweaver-Burk plot: intercept on the $1/V$ axis is $1/V_{max}$; intercept on the $1/[S]$ axis is $-1/K_m$.

Michaelis-Menten equation: $V = \dfrac{V_{max}[S]}{K_m + [S]}$
Double reciprocal form: $\dfrac{1}{V} = \dfrac{K_m}{V_{max}}\cdot\dfrac{1}{[S]} + \dfrac{1}{V_{max}}$
Substrate for a given fraction of $V_{max}$: $[S] = K_m\,\dfrac{f}{1-f}$

Common traps

$K_m$ is not the substrate concentration needed to reach $V_{max}$; it is the concentration that gives half of $V_{max}$.
Changing the amount of enzyme changes $V_{max}$ but not $K_m$.
Cooling does not destroy an enzyme the way heating does, so activity returns on warming.

Shortcuts

Fraction of Vmax from a multiple of Km. Use $[S] = K_m f/(1-f)$ or its inverse $f = [S]/(K_m + [S])$. So $[S] = K_m$ gives 50%, $3K_m$ gives 75%, $4K_m$ gives 80%, $9K_m$ gives 90% and $99K_m$ gives 99% of $V_{max}$; $[S] = K_m/9$ gives 10%.

Temperature coefficient. Below the optimum the rate roughly doubles for each $10^{\circ}$C rise ($Q_{10} \approx 2$): a $20^{\circ}$C rise gives about $4\times$ the rate, valid only while the enzyme remains stable.

Enzyme inhibition and regulation

An inhibitor lowers the rate of an enzyme-catalysed reaction. In competitive inhibition the inhibitor resembles the substrate and competes for the active site; it forms an EI complex but no product. A high substrate concentration crowds out the inhibitor, so $V_{max}$ is unchanged but more substrate is needed to reach half of it, so $K_m$ appears larger. Malonate inhibits succinate dehydrogenase because it resembles succinate; sulpha drugs inhibit a bacterial enzyme by competing with the substrate para-aminobenzoic acid; ethanol is given in methanol poisoning because it competes for alcohol dehydrogenase; statins compete at HMG-CoA reductase.

A non-competitive inhibitor binds at a different site and reduces the amount of active enzyme, whatever the substrate concentration. Substrate cannot outcompete it, so $V_{max}$ falls, while the affinity of the remaining active enzyme for the substrate is unchanged, so $K_m$ is the same. An uncompetitive inhibitor binds only to the ES complex and lowers both $V_{max}$ and $K_m$. Irreversible inhibitors, such as heavy metal ions on -SH groups or nerve agents on a serine at the active site, bind very tightly or covalently. Cyanide is a tight-binding inhibitor of a different kind: it attaches by coordination to the iron of cytochrome c oxidase, but the binding is not covalent.

Cells also regulate their own enzymes. In feedback (end-product) inhibition, the final product of a pathway inhibits an early enzyme of that pathway, so the cell stops making more than it needs: isoleucine, the end product of a pathway from threonine, inhibits threonine deaminase, the first enzyme of that pathway. Product inhibition is related but simpler: the immediate product of a reaction slows the enzyme that makes it, as glucose-6-phosphate does to hexokinase. Many regulated enzymes are allosteric: the regulator binds a separate site and changes the shape of the active site. Allosteric enzymes with several subunits often show a sigmoid, not hyperbolic, curve.

Key points

Competitive: structural analogue of the substrate; binds the active site; $V_{max}$ unchanged, apparent $K_m$ increases; reversed by more substrate.
Non-competitive: binds elsewhere; $V_{max}$ decreases, $K_m$ unchanged; not reversed by more substrate.
Uncompetitive: binds only the ES complex; both $V_{max}$ and $K_m$ decrease.
Examples: malonate (succinate dehydrogenase), sulpha drugs (PABA), ethanol (methanol poisoning), statins.
Feedback inhibition: the final product of a pathway inhibits an early enzyme (isoleucine on threonine deaminase). Product inhibition: a reaction's own product slows it (glucose-6-phosphate on hexokinase). Allosteric regulators bind a site separate from the active site.

Apparent $K_m$ with a competitive inhibitor: $K_m^{app} = K_m\left(1 + \dfrac{[I]}{K_i}\right)$

Common traps

A competitive inhibitor does not lower $V_{max}$; it appears to lower the speed only at modest substrate concentrations.
Adding more substrate does not relieve non-competitive inhibition.
Heating inactivates an enzyme by denaturation, which is not inhibition by a molecule.

Shortcuts

Reading inhibition on a Lineweaver-Burk plot. Compare intercepts. Lines meeting on the $1/V$ axis (same $V_{max}$, different $K_m$) mean competitive inhibition. Lines meeting on the $1/[S]$ axis (same $K_m$, different $V_{max}$) mean non-competitive inhibition. Parallel lines (both intercepts shift in the same ratio) mean uncompetitive inhibition.

Classification and nomenclature of enzymes

The international system sorts enzymes into six main classes by the type of reaction they catalyse, and each enzyme has a four-number Enzyme Commission (EC) code whose first number is the class. Oxidoreductases (class 1) catalyse oxidation-reduction between two substrates (dehydrogenases, oxidases). Transferases (2) move a group other than hydrogen from one substrate to another (transaminases, kinases). Hydrolases (3) split bonds by adding water (amylase, lipase, peptidases, phosphatases).

Lyases (4) remove groups from a substrate by a route other than hydrolysis, leaving a double bond (decarboxylases, aldolase), or add groups to a double bond in the reverse direction. Isomerases (5) rearrange atoms within a molecule (mutases, racemases, epimerases). Ligases (6) join two molecules with a new bond, using energy from ATP, as in the formation of C-O, C-S, C-N or C-C bonds; DNA ligase is the standard example. Since 2018 the international list also has a seventh class, translocases (movers of ions and molecules across membranes, EC 7), but the NEET syllabus teaches six.

A useful naming difference is between synthases and synthetases. By the usual convention a synthase builds a molecule without direct use of ATP (it is typically a lyase or transferase), whereas a synthetase uses ATP to join two substrates (a ligase); a few older names, such as ATP synthase, do not follow the rule. The reaction type does not depend on the substrate; an enzyme that removes $\ce{CO2}$ from a keto acid is a lyase whether it acts on pyruvate or on oxaloacetate.

Key points

1 Oxidoreductases: electron or hydrogen transfer (redox). 2 Transferases: transfer of a group. 3 Hydrolases: cleavage by water.
4 Lyases: cleavage without water, leaving double bonds. 5 Isomerases: interconversion of isomers. 6 Ligases: joining with ATP.
EC number has four parts; the first gives the class.
Ligase: DNA ligase, aminoacyl-tRNA synthetase. Lyase: aldolase, decarboxylase. Isomerase: phosphoglucomutase.
Usually, synthase does not need ATP directly; synthetase does.

Common traps

Hydrolases break bonds with water; lyases break bonds without water. A decarboxylase is a lyase, not a hydrolase.
Kinases transfer a phosphate group from ATP, so they are transferases, not ligases.
Dehydrogenases are oxidoreductases; they do not belong to the hydrolase class because their names resemble 'hydro'.

Shortcuts

Class order mnemonic. The six classes in number order are Oxidoreductases, Transferases, Hydrolases, Lyases, Isomerases, Ligases: first letters O T H L I L, remembered as 'Old Tigers Hunt Lazy Idle Lambs'. The class number is the first digit of the EC code, so EC 3.x.x.x is a hydrolase and EC 6.x.x.x is a ligase.

Synthase versus synthetase. Does the name mean ATP is used? Usually yes for synthetase (a ligase, class 6), and usually no for synthase (a lyase or transferase); a few older names are exceptions.

Cofactors, coenzymes and prosthetic groups

Many enzymes need a non-protein helper to work. The protein part alone is the apoenzyme and is inactive; adding the cofactor gives the active holoenzyme. Removing the cofactor, for example by dialysis, removes the activity. Cofactors fall into three groups. Prosthetic groups are organic molecules bound tightly (sometimes covalently) to the enzyme throughout the reaction; haem in peroxidase and catalase is the textbook example. Coenzymes are organic molecules that associate with the enzyme only during the reaction and are released afterwards; many are derived from vitamins. Metal ions bind to side chains at the active site (and often to the substrate) through coordinate bonds, as zinc does in carboxypeptidase and carbonic anhydrase.

Coenzymes act as carriers that move electrons, hydrogen atoms or chemical groups between enzymes. NAD and NADP, derived from niacin (vitamin B3), carry hydride in redox reactions; FAD (from riboflavin, B2) carries hydrogen; coenzyme A (from pantothenic acid, B5) carries acyl groups; thiamine pyrophosphate (from B1) assists decarboxylations; pyridoxal phosphate (B6) works in amino acid transformations; biotin (B7) carries carbon dioxide in carboxylases. Because many coenzymes derive from vitamins, a vitamin deficiency may cut the activity of the enzymes that need them.

Key points

Apoenzyme (protein only) + cofactor = holoenzyme (active).
Prosthetic group: tightly bound organic part (haem in catalase, peroxidase). Coenzyme: loosely and temporarily bound organic carrier (NAD, NADP, FAD, CoA).
Metal ions: zinc (carboxypeptidase, carbonic anhydrase, alcohol dehydrogenase), iron, copper, magnesium, manganese, molybdenum.
Many coenzymes are derived from vitamins.
Removing a required cofactor abolishes catalysis.

Common traps

A metal ion cofactor is not a coenzyme: the word coenzyme is reserved for organic cofactors.
The apoenzyme alone is inactive; the activity belongs to the holoenzyme.
A vitamin is not itself the active coenzyme in most cases; it is converted into the coenzyme form inside the cell.

Shortcuts

Vitamin to coenzyme. B1 gives thiamine pyrophosphate (TPP); B2 gives FAD and FMN; B3 (niacin) gives NAD and NADP; B5 (pantothenic acid) gives coenzyme A; B6 gives pyridoxal phosphate; B7 (biotin) is the carboxylase cofactor.

Metabolism and the living state

Metabolism is the whole network of chemical reactions in a cell. A metabolic pathway is a series of reactions in which each product is the substrate of the next; glycolysis, which converts glucose to pyruvate in ten steps, is the standard example. Reactions that build large molecules from small ones, such as making protein from amino acids, are anabolic and need energy. Reactions that break larger molecules down, such as the oxidation of glucose, are catabolic and release energy, which is captured partly in ATP. Cells couple the two: the energy released by catabolism is used to drive anabolic reactions that would not otherwise go forward.

Enzymes catalyse nearly all of these reactions, and the core pathways are much the same in every organism from bacteria to humans, one of the clearest signs of the unity of life. A system at chemical equilibrium has no net reactions and can do no work, so a cell at equilibrium would be dead. A living cell stays away from equilibrium because it is open: nutrients flow in, wastes flow out, and the concentration of each compound holds steady because the rate of supply equals the rate of use. This steady state needs a constant energy supply, and metabolism is what provides it.

Key points

Metabolism: anabolism (synthesis, energy-requiring) + catabolism (breakdown, energy-releasing).
Metabolic pathway: sequence of enzyme-catalysed reactions with intermediates; glycolysis has 10 steps.
ATP couples energy-releasing to energy-requiring reactions.
A living organism is an open system at steady state, not at equilibrium; equilibrium means death.
The basic metabolic pathways are shared by all forms of life.

Common traps

A steady state is not an equilibrium: concentrations stay constant because inflow equals outflow while energy is supplied.
Anabolism is not defined by the production of heat or ATP; it is the building of larger molecules and consumes energy.

Primary and secondary metabolites

Tissue analysis finds thousands of different small molecules. Primary metabolites are those with identifiable functions in growth, development and reproduction, and so in normal physiology: amino acids, sugars, nucleotides, fatty acids and lipids. Every cell makes them.

Secondary metabolites are compounds that are not part of the basic growth machinery of the cell but are made in large amounts in particular organisms. Their roles are often ecological, as defence or attraction, and many are of great use to humans. They fall into chemical families. Alkaloids include morphine and codeine; terpenoids give essential oils and many pigments; carotenoids and anthocyanins colour flowers and fruit; abrin and ricin are plant toxins; lectins such as concanavalin A bind sugars; vinblastine and curcumin are used as drugs; and rubber and gums are polymeric. School classification tables also place cellulose in the polymer group. Antibiotics from microbes are secondary metabolites too.

Key points

Primary metabolites: amino acids, sugars, nucleotides, lipids; known roles in normal growth and physiology.
Secondary metabolites: alkaloids, terpenoids, essential oils, pigments, toxins, lectins, drugs, rubber and gums.
Secondary metabolites often have ecological significance and many human uses (drugs, spices, colours, scents).

Common traps

The school classification of secondary metabolites includes cellulose under polymeric substances, although cellulose is a structural carbohydrate.
Primary versus secondary is a division by role in basic physiology, not by size or by chemical class.

Worked examples

Finding Km from a single velocity measurement

An enzyme has $V_{max} = 80\ \mu\text{mol min}^{-1}$. At $[S] = 2\ \text{mM}$ the measured velocity is $20\ \mu\text{mol min}^{-1}$. Find $K_m$, and then the velocity at $[S] = 18\ \text{mM}$.

Write the equation. $V = \dfrac{V_{max}[S]}{K_m + [S]}$. Substituting gives $20 = \dfrac{80 \times 2}{K_m + 2}$.
Solve for Km. $K_m + 2 = \dfrac{160}{20} = 8$, so $K_m = 6\ \text{mM}$. Check: $V$ at 2 mM is a quarter of $V_{max}$, which is the value expected when $[S] = K_m/3$ (since $f = 1/(1+3) = 0.25$).
Velocity at 18 mM. $18 = 3K_m$, so $f = 3/(1+3) = 0.75$ and $V = 0.75 \times 80 = 60\ \mu\text{mol min}^{-1}$. Direct calculation: $80 \times 18/(6 + 18) = 1440/24 = 60$.
Effect of doubling the enzyme. Doubling $[E]_{total}$ doubles $V_{max}$ to $160$ but leaves $K_m = 6\ \text{mM}$; the velocity at 18 mM would then be $120$.

Answer: $K_m = 6\ \text{mM}$ and $V(18\ \text{mM}) = 60\ \mu\text{mol min}^{-1}$.

A competitive inhibitor: how much extra substrate restores the rate?

An enzyme has $K_m = 2\ \text{mM}$ and $V_{max} = 100$ (arbitrary units). A competitive inhibitor with $K_i = 2\ \text{mM}$ is present at $[I] = 6\ \text{mM}$. (a) Find the apparent $K_m$. (b) Compare the velocity at $[S] = 8\ \text{mM}$ with and without inhibitor. (c) What $[S]$ gives back the uninhibited velocity at 8 mM?

Apparent Km. $K_m^{app} = K_m(1 + [I]/K_i) = 2(1 + 6/2) = 8\ \text{mM}$. $V_{max}$ stays $100$.
Velocities at 8 mM. Without inhibitor: $V = 100 \times 8/(2 + 8) = 80$. With inhibitor: $V = 100 \times 8/(8 + 8) = 50$. The inhibitor lowers the velocity by $37.5\%$ at this substrate level.
Restoring 80. Need $\dfrac{[S]}{8 + [S]} = 0.8$, so $[S] = 4 \times 8 = 32\ \text{mM}$ (using $[S] = K_m^{app} f/(1-f)$ with $f = 0.8$). Check: $100 \times 32/(8 + 32) = 80$.
Conclusion about Vmax. Because $V_{max}$ is unchanged, enough substrate always restores the velocity. A non-competitive inhibitor would not allow this.

Answer: (a) $8\ \text{mM}$; (b) $80$ without, $50$ with the inhibitor; (c) $[S] = 32\ \text{mM}$.

Reading Vmax, Km and the type of inhibition from a double-reciprocal plot

Velocities (in $\mu\text{mol min}^{-1}$) without inhibitor: $20$ at $[S] = 1\ \text{mM}$ and $50$ at $[S] = 4\ \text{mM}$. With an inhibitor: $10$ at $1\ \text{mM}$ and $25$ at $4\ \text{mM}$. Find $V_{max}$ and $K_m$ in each case and name the type of inhibition.

Set up the line without inhibitor. Points $(1/[S], 1/V) = (1, 0.05)$ and $(0.25, 0.02)$. Slope $= (0.05 - 0.02)/(1 - 0.25) = 0.04$; intercept $= 0.05 - 0.04 \times 1 = 0.01$.
Convert. $V_{max} = 1/0.01 = 100$. $K_m = \text{slope} \times V_{max} = 0.04 \times 100 = 4\ \text{mM}$. Check: $100 \times 1/(4 + 1) = 20$ and $100 \times 4/8 = 50$.
The inhibited line. Points $(1, 0.10)$ and $(0.25, 0.04)$. Slope $= 0.06/0.75 = 0.08$; intercept $= 0.10 - 0.08 = 0.02$. So $V_{max} = 50$ and $K_m = 0.08 \times 50 = 4\ \text{mM}$.
Classify. $K_m$ is unchanged and $V_{max}$ is halved; the two lines meet on the $1/[S]$ axis at $-1/K_m = -0.25$. This is non-competitive inhibition.

Answer: Without inhibitor: $V_{max} = 100$, $K_m = 4\ \text{mM}$. With inhibitor: $V_{max} = 50$, $K_m = 4\ \text{mM}$; non-competitive inhibition.

Bonds, water, mass and sequences of a tripeptide

A linear tripeptide is made from glycine ($75.07\ \text{g/mol}$), alanine ($89.09$) and serine ($105.09$), one of each. (a) How many peptide bonds form and how many water molecules are lost? (b) Estimate its molar mass. (c) How many different tripeptides can be built from these three, each used once? (d) How many tripeptides are possible from all 20 amino acids if repeats are allowed?

Bonds and water. Three residues give $3 - 1 = 2$ peptide bonds, and each bond releases one $\ce{H2O}$, so 2 water molecules are lost. The chain has one free $\ce{NH2}$ and one free $\ce{COOH}$.
Mass. $M = 75.07 + 89.09 + 105.09 - 2 \times 18.02 = 269.25 - 36.04 = 233.21\ \text{g/mol}$.
Orderings. Three different amino acids can be arranged in $3! = 6$ orders (for example Gly-Ala-Ser and Ser-Ala-Gly are different molecules, because the N-to-C direction matters).
All sequences. With 20 choices at each of three positions there are $20^3 = 8000$ tripeptides. A tetrapeptide would have $20^4 = 160000$.

Answer: 2 peptide bonds, 2 water lost; about $233.2\ \text{g/mol}$; 6 tripeptides from the three given; 8000 possible tripeptides in all.

Base composition, hydrogen bonds and length of a DNA segment

A double-stranded DNA segment of 1000 base pairs has 30% adenine. Find the percentage of the other three bases, the number of hydrogen bonds holding the strands together, and the length of the segment in B-form (rise $0.34\ \text{nm}$ per base pair, 10 base pairs per turn).

Percentages. A pairs with T, so T $= 30\%$. $A + T = 60\%$, leaving $40\%$ for G and C, which are equal: G = C $= 20\%$.
Counts. The segment has $2 \times 1000 = 2000$ nucleotides, so A = T = 600 and G = C = 400. In base pairs: $600$ A-T pairs and $400$ G-C pairs.
Hydrogen bonds. $H = 2 \times 600 + 3 \times 400 = 1200 + 1200 = 2400$.
Length and turns. Length $= 1000 \times 0.34 = 340\ \text{nm}$. Turns $= 1000/10 = 100$, which agrees with the pitch of $3.4\ \text{nm}$ per turn ($100 \times 3.4 = 340$).

Answer: T 30%, G 20%, C 20%; 2400 hydrogen bonds; length $340\ \text{nm}$ (100 turns).

Why sucrose gives a Benedict's test only after acid hydrolysis

A sucrose solution gives no colour change with Benedict's reagent. After boiling with a little dilute acid and neutralising, it gives a brick-red precipitate. Explain.

What Benedict's reagent detects. It detects sugars that have a free anomeric carbon, one that can open its ring to expose an aldehyde or a ketol group that reduces $\ce{Cu^2+}$ to $\ce{Cu+}$. Such sugars are reducing sugars.
Where the bond is in sucrose. The glycosidic bond of sucrose joins carbon 1 of glucose to carbon 2 of fructose; these are the two anomeric carbons. With both tied up, neither ring can open, so sucrose is non-reducing.
What acid does. Hot dilute acid hydrolyses the glycosidic bond: sucrose + $\ce{H2O}$ $\rightarrow$ glucose + fructose (the enzyme invertase does the same at body temperature).
Why the test now works. Each product has a free anomeric carbon and is a reducing sugar, so the reagent is reduced and the red precipitate appears. The mixture rotates polarised light to the left, so it is called invert sugar.

Answer: Sucrose is non-reducing because both anomeric carbons are in the glycosidic link. Hydrolysis frees them, giving reducing glucose and fructose (invert sugar).

Figure: Skeletal formulas of a general amino acid in neutral form and as a zwitterion: a central alpha carbon bonded to H, an amino group, a carboxyl group and an R side chain, with the amino group as NH3+ and the carboxyl as COO- in the zwitterion.

Figure: Two amino acids with the OH of one carboxyl and an H of the other amino group circled; an arrow leads to a dipeptide in which the carbonyl carbon is bonded to nitrogen (the peptide bond, outlined in red) and water is released.

Figure: Four panels: a chain of coloured beads for primary structure, a helix and three antiparallel arrows for secondary structure, one folded chain with a disulfide bridge for tertiary structure, and four subunits labelled alpha and beta, each with a red haem, for quaternary structure.

Figure: Three stages: an enzyme with a notch-shaped active site and a substrate above it; the substrate bound in the active site as the enzyme-substrate complex; the enzyme unchanged with two products leaving.

Figure: Graph of reaction velocity against substrate concentration: a hyperbola rising steeply and then levelling toward a dashed line marked Vmax, with half of Vmax and the matching substrate concentration Km marked by dotted lines.

Figure: Double-reciprocal plot with three straight lines: no inhibitor, competitive inhibitor and non-competitive inhibitor. The competitive line meets the no-inhibitor line on the 1/V axis; the non-competitive line meets it on the 1/[S] axis at the same negative intercept.

Figure: Energy profile from substrate S to product P with two humps: a high red curve for the uncatalysed reaction and a lower blue curve for the enzyme-catalysed reaction; vertical arrows mark the activation energies and both curves start and end at the same levels.

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