Chemistry of Biomolecules: Structure, Sequencing, Purification and Spectroscopy
2. Nucleotides and nucleic acids
A nucleotide has three parts: a nitrogenous base, a pentose and one or more phosphate groups. The purines adenine and guanine and the pyrimidines cytosine, thymine (DNA) and uracil (RNA) are attached by a β-N-glycosidic bond from C1′ of the sugar to N9 of a purine or N1 of a pyrimidine. Base plus sugar is a nucleoside (adenosine); its phosphate ester at C5′ is a nucleotide (AMP, ATP). RNA contains ribose and DNA 2′-deoxyribose. Nucleotides are linked by 3′→5′ phosphodiester bonds into a sugar–phosphate backbone with a 5′-phosphate end and a 3′-OH end, and sequences are written 5′→3′. In alkali the 2′-OH of RNA attacks the neighbouring phosphodiester, forming a 2′,3′-cyclic phosphate and breaking the chain, so RNA is alkali-labile while DNA is not.
In the B-DNA double helix two antiparallel strands wind into a right-handed helix with about 10.5 base pairs per turn and a rise of 0.34 nm per base pair. A pairs with T through two hydrogen bonds and G with C through three, which gives Chargaff’s rules for double-stranded DNA: A = T, G = C, and purines = pyrimidines. A duplex with 30% A therefore has 30% T, and the remaining 40% splits into 20% G and 20% C. Stacked bases absorb less ultraviolet light than free ones, so melting a duplex raises its A260 (the hyperchromic effect); the melting temperature Tm rises with GC content and with ionic strength.
3. Peptide sequencing by chemical and enzymatic methods
Chemical sequencing of a protein follows a fixed plan.
(1) Break the disulfides, by oxidation with performic acid (to cysteic acid) or by reduction with 2-mercaptoethanol or dithiothreitol followed by blocking the thiols with iodoacetate.
(2) Find the amino acid composition by hydrolysis in 6 M HCl at 110 °C for about 24 h; Trp is destroyed, and Asn and Gln are converted to Asp and Glu.
(3) Identify the ends: the N-terminal residue with 1-fluoro-2,4-dinitrobenzene (Sanger’s reagent, giving a yellow DNP-amino acid) or dansyl chloride (giving a fluorescent derivative), both followed by total hydrolysis; the C-terminal residue with carboxypeptidases or by hydrazinolysis.
(4) Cut the chain into fragments and sequence each.
(5) Order the fragments using the overlaps from a second, different cleavage.
Edman degradation removes one residue at a time without hydrolysing the rest. Phenyl isothiocyanate (PITC) couples to the free N-terminal amino group under mildly alkaline conditions; anhydrous trifluoroacetic acid then cleaves off the N-terminal residue as an anilinothiazolinone, which is converted to a stable phenylthiohydantoin (PTH) amino acid and identified by chromatography, while the shortened peptide enters the next cycle. No cycle is 100% efficient, so out-of-step chains accumulate: at 98% yield per cycle only 0.98⁵⁰ ≈ 36% of chains are still in step after 50 cycles. A run therefore reads a few tens of residues, and long proteins must first be cut.
| Agent | Kind | Cleaves | Remark |
|---|---|---|---|
| Trypsin | Enzyme (endopeptidase) | Carboxyl side of Lys and Arg | Not when the next residue is Pro |
| Chymotrypsin | Enzyme (endopeptidase) | Carboxyl side of Phe, Tyr and Trp | Not when the next residue is Pro |
| Cyanogen bromide (CNBr) | Chemical | Carboxyl side of Met | Met becomes C-terminal homoserine lactone |
| Carboxypeptidase A or B | Enzyme (exopeptidase) | Releases C-terminal residues one by one | A spares C-terminal Arg, Lys and Pro; B removes Arg and Lys |
Worked example. A heptapeptide gives PTH-Leu in the first Edman cycle. Trypsin gives Leu-Ala-Lys, Gly-Met-Arg and free Tyr; CNBr gives a pentapeptide ending in homoserine lactone and the dipeptide Arg-Tyr. Leu-Ala-Lys carries the N-terminus and comes first, and Arg-Tyr shows that Tyr follows the Arg of Gly-Met-Arg, so the sequence is Leu-Ala-Lys-Gly-Met-Arg-Tyr.
4. DNA sequencing by chemical and enzymatic methods
Chemical (Maxam–Gilbert) method. A DNA fragment is labelled at one end with ³²P (at the 5′ end, for example, with polynucleotide kinase) and divided into four portions. Each is treated with a reagent that modifies particular bases, under conditions that give about one modification per molecule: dimethyl sulfate methylates G; formic acid attacks the purines A and G; hydrazine attacks C and T; and hydrazine in high salt attacks C only. Hot piperidine then cleaves the backbone at each modified base, and a denaturing gel with autoradiography shows only the labelled fragments. Reading upward from the shortest: G gives a band in the G lane, A in A+G but not G, C in both C and C+T, and T in C+T only.
Enzymatic (Sanger dideoxy, chain-termination) method. A single-stranded template, a primer, DNA polymerase and all four dNTPs are supplied together with a small proportion of one 2′,3′-dideoxynucleoside triphosphate (ddNTP). A ddNTP is incorporated normally but lacks the 3′-OH, so no further nucleotide can be joined and the chain stops. Each reaction therefore holds a nested set of fragments, all starting at the primer and ending at every position of that base. Run in four lanes, or in automated sequencing with four ddNTPs carrying different fluorescent dyes in one capillary, the fragments read from shortest to longest give the newly synthesised strand, 5′→3′; the template is its antiparallel complement. Worked example: if the products, from shortest to longest, end in ddG, ddA, ddT, ddC and ddC, the new strand is 5′-GATCC-3′ and the template region is 5′-GGATC-3′.
5. Carbohydrates up to hexoses
Monosaccharides are polyhydroxy aldehydes (aldoses) or ketones (ketoses), (CH₂O)ₙ. Glyceraldehyde and dihydroxyacetone are the trioses, ribose an aldopentose, glucose, mannose and galactose aldohexoses, and fructose a ketohexose. An aldohexose has four chiral centres (C2 to C5) and so 2⁴ = 16 stereoisomers, eight D and eight L; a ketohexose has three and so eight. D or L refers to the chiral carbon farthest from the carbonyl (C5 in a hexose), compared with D-glyceraldehyde; it does not predict the sign of rotation (D-fructose is laevorotatory). Sugars differing at only one chiral centre are epimers: glucose and mannose at C2, glucose and galactose at C4.
In solution pentoses and hexoses cyclise as a ring OH adds to the carbonyl, giving a hemiacetal (from an aldose) or a hemiketal (from a ketose). Glucose closes C1 with the C5-OH into a six-membered pyranose; fructose closes C2 with the C5-OH into a five-membered furanose (or with the C6-OH into a pyranose). The former carbonyl carbon becomes a new chiral centre, the anomeric carbon, giving α and β anomers. A freshly dissolved pure anomer slowly changes its optical rotation as the ring opens and recloses, until an equilibrium mixture forms: mutarotation. Taking [α]D = +112.2° for α-D-glucopyranose, +18.7° for β and +52.7° at equilibrium, the α fraction x satisfies 112.2x + 18.7(1 − x) = 52.7, so x = 0.36: about 36% α and 64% β.
Every monosaccharide is a reducing sugar: its open-chain aldehyde (for a ketose, the aldehyde formed through an enediol in alkali) reduces Cu²⁺ in Fehling’s or Benedict’s reagent to red Cu₂O. When the anomeric OH is replaced by an O-alkyl group, as in methyl α-D-glucopyranoside, the product is an acetal (glycoside): it cannot open, does not mutarotate and is non-reducing. Bromine water oxidises glucose to gluconic acid (an aldonic acid), nitric acid to glucaric acid, and reduction gives sorbitol (an alditol). Glucose, mannose and fructose give the same osazone, because only C1 and C2 react and they differ only there.
6. Lipids: triglycerides
A triglyceride (triacylglycerol) is the triester of glycerol with three fatty acids: simple if all three are the same (tristearin), mixed otherwise. Common fatty acids are unbranched with an even number of carbons: palmitic (16:0), stearic (18:0), oleic (18:1, cis-Δ9) and linoleic (18:2, cis-Δ9,12). Each cis double bond kinks the chain and prevents close packing, lowering the melting point, so fats rich in saturated chains are solid at room temperature and oils rich in unsaturated chains are liquid; catalytic hydrogenation over nickel hardens oils. Triglycerides are neutral and non-polar, suited to storing energy rather than to forming membranes.
Saponification is alkaline hydrolysis: triglyceride + 3 KOH → glycerol + 3 potassium soaps. The saponification value, the mg of KOH needed to saponify 1 g of fat, is SV = 3 × 56.1 × 1000/M, inversely proportional to the mean molar mass M and so high for short-chain fats. Tristearin (C₅₇H₁₁₀O₆, M = 891.5 g mol⁻¹) has SV = 168,300/891.5 = 189. The iodine value, the grams of I₂ taken up by 100 g of fat, measures unsaturation, one I₂ adding across each C=C. Fats spoil by hydrolytic rancidity (lipases release free fatty acids) and oxidative rancidity (oxygen attacks the double bonds).
7. Purification: ion-exchange and gel-filtration chromatography
Ion-exchange chromatography separates by charge. The matrix carries fixed charged groups with loosely held counter-ions. A cation exchanger carries negative groups, weak carboxymethyl (CM) or strong sulfopropyl (SP); an anion exchanger carries positive groups, weak diethylaminoethyl (DEAE) or strong quaternary ammonium (Q). A protein binds a cation exchanger at a pH below its pI, where it is positive, and an anion exchanger at a pH above its pI. Unbound proteins wash through; bound ones are eluted by raising the salt concentration, whose ions compete for the fixed charges, or by shifting the pH towards their pI, the most highly charged leaving last. Worked example: at pH 7.0 on DEAE, a protein of pI 5.0 (negative) binds, and one of pI 8.5 (positive) passes straight through.
Gel-filtration (size-exclusion) chromatography separates by size and shape on porous beads of cross-linked dextran (Sephadex), agarose or polyacrylamide. Molecules too large to enter the pores travel only in the liquid between the beads and elute first, at the void volume V₀; smaller ones spend time inside the pores and elute later, the smallest near the total volume Vt. For globular proteins Ve falls linearly with log M within the fractionation range, so a calibrated column gives the native molar mass; Kav = (Ve − V₀)/(Vt − V₀) runs from 0 (excluded) to 1 (fully included). Compared with SDS-PAGE, which gives subunit masses, it reveals oligomers.
8. Absorption and fluorescence spectroscopy
Absorbance A = log₁₀(I₀/I) = εcl (the Beer–Lambert law), where ε is the molar absorption coefficient (M⁻¹ cm⁻¹), c the concentration and l the path length, usually 1 cm. Proteins absorb at 280 nm mainly through their Trp and Tyr side chains, with a small contribution from disulfides; a common estimate is ε280 = 5500 nW + 1490 nY + 125 nSS, with n the numbers of Trp, Tyr and disulfide bonds. Peptide bonds absorb at 190 to 220 nm. Nucleic acid bases absorb near 260 nm; an A260/A280 ratio of about 1.8 for DNA (about 2.0 for RNA) is the usual purity check. NADH absorbs at 340 nm (ε = 6220 M⁻¹ cm⁻¹) and NAD⁺ does not, so dehydrogenases are assayed at 340 nm. Worked example: one Trp and two Tyr give ε280 = 5500 + 2 × 1490 = 8480 M⁻¹ cm⁻¹, so A280 = 0.424 in a 1 cm cell means c = 0.424/8480 = 50 µM.
In fluorescence, a molecule raised to an excited singlet state loses some energy by vibrational relaxation and then emits at a longer wavelength than it absorbed (the Stokes shift). Emission is measured at right angles against a dark background, which makes fluorescence far more sensitive than absorbance. Trp dominates protein fluorescence (excitation near 280 nm, or 295 nm to excite Trp selectively); exposed to water it emits near 350 nm, while buried in a non-polar core its emission shifts to shorter wavelengths, so a blue shift reports folding or ligand binding. NADH fluoresces (excitation 340 nm, emission near 460 nm) and NAD⁺ does not. Extrinsic probes add fluorescence where none exists: dansyl derivatives label amino groups, and ethidium bromide fluoresces strongly only when intercalated into DNA.
Key takeaways
- pI is the mean of the two pKa values flanking the neutral form (Gly 5.97, Lys 9.74); below its pI a molecule is positive, above it negative.
- Nucleotides: base to C1′ through N9 (purines) or N1 (pyrimidines), chains linked 3′→5′; A = T and G = C in duplex DNA; RNA is alkali-labile because of its 2′-OH.
- Peptides: Edman’s PITC releases PTH-amino acids cycle by cycle; trypsin cuts after Lys and Arg, chymotrypsin after Phe, Tyr and Trp, CNBr after Met; overlaps order the pieces.
- DNA: Maxam–Gilbert cleaves end-labelled DNA chemically at chosen bases; Sanger’s ddNTPs stop polymerase copies, and the read is the new strand 5′→3′, complementary to the template.
- Monosaccharides are all reducing, glycosides are not; epimers differ at one centre, anomers at the anomeric carbon. A fat’s saponification value falls with molar mass; its iodine value measures unsaturation.
- Ion exchange separates by charge relative to pI, gel filtration by size (largest first). A = εcl at 280 nm (Trp, Tyr), 260 nm (bases) and 340 nm (NADH); a Trp fluorescence blue shift means burial.
Practice questions (18)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
Which of the standard amino acids found in proteins has no chiral carbon atom?
Show answer
Answer: D — Glycine
Glycine’s side chain is a hydrogen atom, so its α-carbon carries two identical H atoms and is not a stereocentre. Alanine, proline and cysteine all have four different groups on the α-carbon; cysteine is the tempting one because it is unusual in being R rather than S, but it is still chiral.Which reagent allows a peptide to be sequenced from its N-terminus one residue per cycle, leaving the rest of the chain intact for the next cycle?
Show answer
Answer: A — Phenyl isothiocyanate, which releases the N-terminal residue as a phenylthiohydantoin derivative
In Edman degradation PITC couples to the N-terminal amino group and anhydrous acid removes only that residue, later converted to a PTH-amino acid, leaving the shortened peptide for the next cycle. FDNB and dansyl chloride also label the N-terminus, but identifying the label requires hydrolysing the whole chain, so only one residue per sample is obtained. CNBr cuts internally at Met and does not read a sequence.An octapeptide gives PTH-Gly in the first Edman cycle, and hydrazinolysis shows Ser at its C-terminus. Trypsin yields three peptides with the compositions (Gly, Tyr, Arg), (Met, Ala, Lys) and (Phe, Ser). Cyanogen bromide yields two peptides, one containing Gly, Tyr, Arg and homoserine lactone and the other Ala, Lys, Phe and Ser. What is the sequence of the octapeptide, written N- to C-terminus in one-letter code (G Gly, Y Tyr, R Arg, M Met, A Ala, K Lys, F Phe, S Ser)?
Show answer
Answer: C — GYRMAKFS
Trypsin cuts after Lys and Arg, so (Gly, Tyr, Arg) ends in Arg and, holding Gly, is N-terminal: Gly-Tyr-Arg. (Phe, Ser) is C-terminal because Ser is the C-terminus: Phe-Ser. The CNBr fragment Gly, Tyr, Arg plus homoserine lactone puts Met directly after Arg, so the middle tryptic peptide is Met-Ala-Lys and the sequence is GYRMAKFS. With Met-Lys-Ala (GYRMKAFS), trypsin would give Met-Lys and Ala-Phe-Ser; ending Ser-Phe contradicts the C-terminal Ser; and swapping Lys and Arg changes the tryptic compositions.In a Sanger dideoxy sequencing reaction primed on a single-stranded template, the terminated products, listed from shortest to longest (each one nucleotide longer than the previous), end in ddC, ddG, ddA, ddT, ddT, ddC, ddA and ddG. What is the sequence of the corresponding region of the template, written 5′→3′?
Show answer
Answer: B — 5′-CTGAATCG-3′
Reading the products from shortest to longest gives the new strand 5′-CGATTCAG-3′. The template pairs with it antiparallel: 3′-GCTAAGTC-5′, which written 5′→3′ is 5′-CTGAATCG-3′. 5′-CGATTCAG-3′ is the new strand itself, 5′-GCTAAGTC-3′ is the complement without reversing its direction, and 5′-GACTTAGC-3′ is the new strand merely reversed.How are D-glucose and D-galactose related to each other?
Show answer
Answer: D — C-4 epimers
D-Glucose and D-galactose differ only in the configuration at C4, so they are C4 epimers. The C2 epimer of glucose is mannose; anomers differ only at the anomeric carbon of the ring form (α and β); and the enantiomer of D-glucose is L-glucose, which differs at every chiral centre.Which of the following does NOT reduce Benedict’s reagent when heated with it?
Show answer
Answer: A — Methyl α-D-glucoside
In methyl α-D-glucoside the anomeric carbon is an acetal: it cannot open to a free aldehyde, so it neither mutarotates nor reduces Cu²⁺. Fructose is the tempting distractor because it is a ketone, but in hot alkali it isomerises through an enediol to aldoses and reduces Benedict’s reagent. Glyceraldehyde and 2-deoxyribose have free aldehyde groups.Three proteins, X (pI 4.8), Y (pI 7.6) and Z (pI 9.8), are loaded in a low-salt buffer at pH 6.0 onto a carboxymethyl (CM) cellulose column, which is then washed and eluted with a steadily increasing NaCl concentration. Assume that a protein’s net charge, and hence its binding strength, grows with the distance between its pI and the buffer pH. In what order do the proteins leave the column?
Show answer
Answer: A — X, then Y, then Z
CM-cellulose is a cation exchanger (fixed –CH₂COO⁻ groups). At pH 6.0, X (pI 4.8) is negative and does not bind, so it leaves in the wash. Y (pI 7.6) and Z (pI 9.8) are positive and bind; Y, 1.6 units from its pI, is released at lower salt than Z, 3.8 units away. The reverse order is what one would expect by mistaking CM for an anion exchanger.A purified enzyme elutes from a calibrated gel-filtration column at the position expected for a 120 kDa globular protein, whereas SDS-PAGE in the presence of 2-mercaptoethanol shows a single band at 60 kDa. What is the most reasonable interpretation?
Show answer
Answer: B — It is a dimer of 60 kDa subunits, which SDS and the thiol reagent separate
Gel filtration measures the native assembly (about 120 kDa), while SDS with a reducing agent unfolds the protein and separates its polypeptides, non-covalent or disulfide-linked, which run at 60 kDa. Two 60 kDa chains make a dimer. Gel filtration separates by size, not charge; SDS denatures but does not hydrolyse peptide bonds; and 30 kDa subunits would run at 30 kDa, not 60.A denatured protein is allowed to refold. Its tryptophan fluorescence emission maximum, with excitation at 295 nm, shifts from about 350 nm to about 330 nm. What does this shift most directly indicate?
Show answer
Answer: C — Its tryptophan side chains have moved into a less polar, buried environment
Trp emission is sensitive to polarity: in water it peaks near 350 nm, and in a non-polar protein interior it shifts to shorter wavelengths, so a blue shift on refolding means the Trp residues have been buried. More exposure would shift emission to longer wavelengths, excitation at 295 nm hardly excites Tyr, and a change in absorbance alters intensity, not the position of the emission maximum.Which of the following statements about nucleotides and nucleic acids are correct?
Show answer
Answer: B — RNA but not DNA is hydrolysed by dilute alkali, because of its 2′-OH groups; C — In double-stranded DNA the mole fraction of purines equals that of pyrimidines
The 2′-OH of RNA attacks the adjacent phosphodiester in alkali (via a 2′,3′-cyclic phosphate), which DNA lacks; and since every A pairs with T and every G with C, purines equal pyrimidines in a duplex. Pyrimidines are linked through N1 (N9 is the purine position), and melting raises A260, the hyperchromic effect, because unstacked bases absorb more.Which of the following statements about DNA sequencing methods are correct?
Show answer
Answer: A — In Sanger’s method a chain stops after a ddNTP is added, because the ddNTP lacks the 3′-OH needed for the next phosphodiester bond; D — In the Maxam–Gilbert method dimethyl sulfate modifies guanine, and piperidine cleaves the backbone at modified bases
A dideoxynucleotide has no 3′-OH, so polymerase cannot add the next nucleotide; and in the chemical method DMS methylates G, after which piperidine breaks the strand there. Maxam–Gilbert uses no polymerase or primer, only base-specific chemistry on end-labelled DNA. In Sanger’s method synthesis runs 5′→3′ from the primer, so the shortest fragments give the 5′ end of the new strand.Which of the following statements about ion-exchange and gel-filtration chromatography are correct?
Show answer
Answer: C — In gel filtration, a protein too large to enter the pores elutes in the void volume, ahead of smaller ones; D — A protein binds to a DEAE (anion) exchanger at a buffer pH above its pI
Excluded molecules travel only outside the beads and leave first, at V₀; and above its pI a protein is negative and binds the positively charged DEAE groups. Bound proteins are eluted by raising, not lowering, the salt concentration, since the salt ions compete for the fixed charges. Gel filtration separates by size and shape only, so it cannot resolve proteins that differ only in charge.Which of the following statements about absorption and fluorescence spectroscopy of biomolecules are correct?
Show answer
Answer: A — Fluorescence emission occurs at a longer wavelength than the light absorbed; C — At a fixed wavelength and concentration, doubling the path length doubles the absorbance
Energy is lost by vibrational relaxation before emission, so fluorescence appears at longer wavelength (the Stokes shift), and A = εcl is proportional to l. Only the reduced NADH absorbs at 340 nm, which is why dehydrogenase assays work there. Protein absorbance at 280 nm comes from Trp and Tyr side chains; peptide bonds absorb at 190 to 220 nm.Lysine has pKa values of 2.18 (α-COOH), 8.95 (α-NH₃⁺) and 10.53 (side-chain ε-NH₃⁺). Its isoelectric point is ____ (to two decimal places).
Numerical answer — type the value.
Show answer
Answer: 9.74
Fully protonated lysine has charge +2; losing the α-COOH proton (2.18) gives +1, losing the α-NH₃⁺ proton (8.95) gives the neutral form, and losing the ε-NH₃⁺ proton (10.53) gives −1. The neutral form lies between 8.95 and 10.53, so pI = (8.95 + 10.53)/2 = 9.74. Averaging 2.18 and 8.95 (5.57), as for a neutral amino acid, is the trap.A protein contains 3 Trp and 6 Tyr residues and 2 Cys residues joined in one disulfide bond. Take molar absorption coefficients at 280 nm of 5500 M⁻¹ cm⁻¹ per Trp, 1490 M⁻¹ cm⁻¹ per Tyr and 125 M⁻¹ cm⁻¹ per disulfide, and assume they add. A solution of the protein has A280 = 0.767 in a 1.00 cm cuvette. Its concentration is ____ µM (to the nearest integer).
Numerical answer — type the value.
Show answer
Answer: 30
ε280 = 3(5500) + 6(1490) + 1(125) = 16,500 + 8940 + 125 = 25,565 M⁻¹ cm⁻¹. By Beer–Lambert, c = A/(εl) = 0.767/(25,565 × 1.00) = 3.00 × 10⁻⁵ M = 30 µM. Leaving out the Tyr contribution is the usual slip; it gives 0.767/16,625 = 46 µM.Triolein (glyceryl trioleate, M = 885.4 g mol⁻¹) has three C=C double bonds per molecule, each of which adds one molecule of I₂ (M = 253.8 g mol⁻¹). Its iodine value, in grams of I₂ taken up per 100 g of fat, is ____ (to the nearest integer).
Numerical answer — type the value.
Show answer
Answer: 86
One mole of triolein (885.4 g) takes up 3 mol of I₂ = 3 × 253.8 = 761.4 g. Per 100 g of fat: 761.4 × 100/885.4 = 86.0, so the iodine value is 86. Forgetting the factor 3 gives 29; tristearin, with no double bonds, would have an iodine value of 0.Take the specific rotations of α-D-galactopyranose and β-D-galactopyranose as +150.7° and +52.8°, and that of their equilibrium mixture in water as +80.2°. Assuming the open-chain and furanose forms are negligible, the percentage of the α anomer at equilibrium is ____ % (to the nearest integer).
Numerical answer — type the value.
Show answer
Answer: 28
If x is the α fraction, 150.7x + 52.8(1 − x) = 80.2, so 97.9x = 27.4 and x = 0.280, i.e. 28% α and 72% β. Working the fraction from the wrong end, (150.7 − 80.2)/97.9 = 72%, gives the β percentage instead.The peptide Gly-Leu-Lys-Ala-Arg-Pro-Glu-Phe-Lys-Ser-Met-Arg-Trp-Tyr-Lys-Asp-Ala is digested to completion with trypsin. The number of peptides produced is ____.
Numerical answer — type the value.
Show answer
Answer: 5
Trypsin cuts on the carboxyl side of Lys and Arg unless the next residue is Pro. Cuts occur after Lys-3, Lys-9, Arg-12 and Lys-15, but not after Arg-5, which is followed by Pro. Four cuts give five peptides: Gly-Leu-Lys, Ala-Arg-Pro-Glu-Phe-Lys, Ser-Met-Arg, Trp-Tyr-Lys and Asp-Ala. Cutting after the Arg-Pro bond as well would give six.