IChO 1986 · Theory problem 6

Restriction endonucleases Cla I and Taq I and recombinant DNA

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In recombinant DNA technology specific endonucleases can recognize and hydrolyse the phosphoric ester bound in each of both strands. Cla I for example hydrolyses the bond between two nucleotides in the sequence:

Figure 1

6.1 Give the base sequence of the complementary strand in the 5' - 3' direction and indicate with arrows the location where the hydrolysis by Cla I would occur.

6.2 How often on average will this sequence occur in one strand of DNA molecule of \(10^5\) base pairs? You can assume that the four bases occur equally often and that they randomly distribute in the two chains.

Figure 2

Taq I hydrolyses a long double strand DNA molecule into fragments which are on average 256 base pairs long. The 3' end of these fragments treated by cleavage turns out to be a thymine(T)- and the 5' end a cytosine(C)- end.

6.3 How long is the sequence recognized by Taq I?

6.4 Give the two possible base sequences (in the direction 5' - 3') which form the recognition pattern for Taq I (must obviously have some symmetry).

The DNA of a phage which occurs as a close circle contains only 5'-pApTpCpGpApT-3' sequence in each of the two strands. After treatment with ClaI equilibrium is established:

Figure 3

6.5 Give a schematic drawing of the circular and linear molecules. Indicate the bases adjacent to the cleaning site in both strands. Indicate also the 3' and 5' ends.

In Fig. 1 the percentage of linear DNA is given as a function of temperature, measured in a solution of 0.15 M \(\ce{NaCl}\) buffered with citrate at pH = 6.5. With Taq I as cleavage enzyme, the same curve is obtained.

6.6 Is the reaction as written endothermic or exothermic? Explain your answer.

6.7 Show, considering the information given, which of the two base sequences of the answer to 6.4 is the correct one.

6.8 What would look the curve for Taq I like if the recognition pattern would have been the other possibility of d)?

A large DNA molecule is cut into fragments with the aid of Cla I. One fragment is isolated, purified and mixed in the ratio of 1:1 with phage DNA which was also cleaved with Cla I. Thereby recombinant molecules can be formed through the reaction:

Figure 4

6.9 Would the enthalpy of this reaction be positive, negative or about zero? Explain your answer.

6.10 Which combination of temperature, DNA concentration and ionic strength (high or low in each case) will give the maximum percentage of recombinant molecules?

Official English text of the problem.

Answers

  • 6.1: The complementary strand, written \(5'\rightarrow3'\), is also \(5'\text{-AT}\downarrow\text{CGAT-}3'\). Cla I produces two-nucleotide 5'-CG cohesive ends.
  • 6.2: Approximately 24 occurrences in one strand: \(10^5/4^6\approx24.4\).
  • 6.3: Four nucleotides per strand, corresponding to a four-base-pair recognition site.
  • 6.4: \(5'\text{-T}\downarrow\text{CGA-}3'\) and \(5'\text{-GAT}\downarrow\text{C-}3'\).
  • 6.5: Cutting the single site opens the circular duplex into one linear molecule with a 5'-CG overhang at each end, terminal 5'-phosphorylated C residues, and terminal 3'-hydroxyl T residues.
  • 6.6: Circular \(\rightarrow\) linear is endothermic, \(\Delta H>0\).
  • 6.7: Taq I recognizes \(5'\text{-TCGA-}3'\) and cleaves \(T\downarrow C\).
  • 6.8: The alternative would give a similar rising curve shifted toward lower temperatures.
  • 6.9: Negative enthalpy, \(\Delta H<0\), for association of the open, cleaved molecules through their cohesive ends.
  • 6.10: Low temperature, low DNA concentration, and high ionic strength. Dilution suppresses formation of chains containing several DNA molecules; excessive dilution would also suppress the initial phage–fragment association.

Solution

6.1

The base symbols denote residues of adenine, A (Nc1ncnc2[nH]cnc12); thymine, T (Cc1c[nH]c(=O)[nH]c1=O); cytosine, C (Nc1cc[nH]c(=O)n1); and guanine, G (Nc1nc2[nH]cnc2c(=O)[nH]1). The SMILES in this sentence describe the corresponding free nucleobases.

A pairs with T, and C pairs with G. The strands run antiparallel, so the aligned recognition site and cleavage positions are

\[ \begin{array}{rccccccccl} 5'&A&T&\downarrow&C&G&&A&T&3'\\ 3'&T&A&&G&C&\uparrow&T&A&5' \end{array} \]

Reading the lower strand from its own 5' end gives

\[ \boxed{5'\text{-AT}\downarrow\text{CGAT-}3'}. \]

Thus ATCGAT is a reverse-complement palindrome. The staggered cuts leave two-nucleotide 5'-CG overhangs.

Each cleavage hydrolyses a phosphodiester linkage, producing a 3'-hydroxyl end and a 5'-phosphate end. A balanced representation, with \(R\) and \(R'\) standing for the sugar-containing DNA portions, is

\[ \ce{R-O-P(=O)(O^-)-O-R' + H2O -> R-OH + HO-P(=O)(O^-)-O-R'}. \]

This equation uses one possible protonation state of the terminal phosphate; the buffer controls subsequent proton exchange. Each double-strand cut consumes two water molecules. Water has SMILES O; the generic phosphodiester and phosphate monoester groups have SMILES *OP(=O)([O-])O* and *OP(=O)([O-])O, respectively, where * denotes an unspecified DNA portion.

For structural identification, the recognition sequences can be represented as isolated 5'-phosphorylated, 3'-hydroxy oligodeoxynucleotides. Their SMILES below encode the natural sugar stereochemistry and use neutral phosphate acid forms for notation. They do not encode duplex pairing or helicity. Complete SMILES for the phage and recombinant DNA cannot be specified because their remaining sequences are not given.

5'-Phosphorylated DNA hexamer d(ATCGAT):

O=P(O)(O)OC[C@H]1O[C@@H](n2cnc3c(N)ncnc32)C[C@@H]1OP(=O)(O)OC[C@H]1O[C@@H](n2cc(C)c(=O)[nH]c2=O)C[C@@H]1OP(=O)(O)OC[C@H]1O[C@@H](n2ccc(N)nc2=O)C[C@@H]1OP(=O)(O)OC[C@H]1O[C@@H](n2cnc3c(=O)[nH]c(N)nc32)C[C@@H]1OP(=O)(O)OC[C@H]1O[C@@H](n2cnc3c(N)ncnc32)C[C@@H]1OP(=O)(O)OC[C@H]1O[C@@H](n2cc(C)c(=O)[nH]c2=O)C[C@@H]1O

6.2

At a specified starting position, the probability of the particular six-base sequence is

\[ p=\left(\frac14\right)^6=\frac1{4096}. \]

For a linear strand containing \(N=10^5\) nucleotides, there are \(N-6+1\) possible starting positions. Therefore,

\[ \mathbb E[n] =(10^5-5)\frac1{4096} =24.4128\ldots \approx\boxed{24\text{ occurrences}}. \]

Neglecting the five terminal positions gives \(24.4141\), essentially the same result. The mean spacing is approximately \(4096\) base pairs.

There is no factor of two: the question asks about one strand, and the complementary occurrence identifies the same duplex site.

6.3

For a particular recognition sequence of length \(m\), the occurrence probability is \(4^{-m}\). In a long DNA molecule, the mean fragment length is consequently approximately \(4^m\) base pairs.

Thus,

\[ 4^m=256=4^4, \qquad \boxed{m=4}. \]

6.4

A terminal 3'-T and terminal 5'-C mean that cleavage occurs at

\[ 5'\text{-}\cdots T\downarrow C\cdots\text{-}3'. \]

A palindromic four-base recognition site has its first and fourth bases complementary, and its second and third bases complementary.

There are two possible placements of \(T\downarrow C\):

Position of \(T\downarrow C\) Recognition sequence Aligned complementary strand
First–second bases \(5'\text{-T}\downarrow\text{CGA-}3'\) \(3'\text{-AGC}\uparrow\text{T-}5'\)
Third–fourth bases \(5'\text{-GAT}\downarrow\text{C-}3'\) \(3'\text{-C}\uparrow\text{TAG-}5'\)

It cannot occupy the middle two positions because T and C are not complementary.

The two candidates therefore produce different cohesive ends:

  • TCGA: two-nucleotide 5'-CG overhangs.
  • GATC: two-nucleotide 3'-AT overhangs.

5'-Phosphorylated DNA tetramer d(TCGA):

O=P(O)(O)OC[C@H]1O[C@@H](n2cc(C)c(=O)[nH]c2=O)C[C@@H]1OP(=O)(O)OC[C@H]1O[C@@H](n2ccc(N)nc2=O)C[C@@H]1OP(=O)(O)OC[C@H]1O[C@@H](n2cnc3c(=O)[nH]c(N)nc32)C[C@@H]1OP(=O)(O)OC[C@H]1O[C@@H](n2cnc3c(N)ncnc32)C[C@@H]1O

5'-Phosphorylated DNA tetramer d(GATC):

O=P(O)(O)OC[C@H]1O[C@@H](n2cnc3c(=O)[nH]c(N)nc32)C[C@@H]1OP(=O)(O)OC[C@H]1O[C@@H](n2cnc3c(N)ncnc32)C[C@@H]1OP(=O)(O)OC[C@H]1O[C@@H](n2cc(C)c(=O)[nH]c2=O)C[C@@H]1OP(=O)(O)OC[C@H]1O[C@@H](n2ccc(N)nc2=O)C[C@@H]1O

6.5

The requested drawing should show the following.

For the original circular molecule, draw two continuous antiparallel circular strands. At the unique recognition site, place the paired bases and staggered cleavage positions exactly as in 6.1. The intact circle has no free 3' or 5' termini; these labels indicate strand direction.

For the linear molecule, open both backbones at those positions and separate the cohesive ends. Reading the resulting duplex from left to right:

  • The upper strand starts \(5'\text{-pCGAT}\cdots\) and ends \(\cdots\text{AT-}3'\).
  • The lower strand starts \(3'\text{-TA}\cdots\) and ends \(\cdots\text{TAGC-p-}5'\).

At the left end, the upper strand’s initial CG projects beyond the lower strand. At the right end, the lower strand’s terminal GC, read left to right, projects beyond the upper strand. Read in its own \(5'\rightarrow3'\) direction, that overhang is also CG.

Thus every 5' terminus is a phosphorylated C, and every 3' terminus is a T bearing a free hydroxyl group.

After cleavage, these complementary overhangs can pair again to produce a circular, annealed molecule with two backbone nicks. This is the circular species participating in the temperature-dependent equilibrium:

\[ \ce{DNA_{circular,annealed} <=> DNA_{linear}}. \]

Annealing alone does not restore the hydrolysed covalent bonds; covalent sealing would require an additional ligation step.

6.6

The fraction of linear DNA increases as the temperature increases. Heating therefore favours the forward direction,

\[ \ce{DNA_{circular,annealed} -> DNA_{linear}}, \]

so this direction is endothermic:

\[ \boxed{\Delta H>0}. \]

More formally, if \(f\) is the fraction of linear molecules,

\[ K=\frac{[\text{linear DNA}]}{[\text{circular DNA}]} =\frac{f}{1-f}. \]

The graph shows \(K\) increasing with temperature. The van ’t Hoff relation,

\[ \frac{d\ln K}{dT}=\frac{\Delta H^\circ}{RT^2}, \]

then gives \(\Delta H^\circ>0\).

Physically, opening the annealed circle disrupts the base pairing and stacking at its cohesive junction. This conclusion concerns end dissociation, not the initial phosphodiester hydrolysis.

6.7

Cla I produces 5'-CG overhangs. Closing their junction forms two G–C base pairs.

Of the two Taq I candidates:

  • \(T\downarrow CGA\) produces the same 5'-CG overhangs.
  • \(GAT\downarrow C\) produces 3'-AT overhangs, which would form two A–T pairs.

The identical temperature curves therefore select, within the model of the problem,

\[ \boxed{5'\text{-T}\downarrow\text{CGA-}3'}. \]

Indeed, this recognition sequence lies within the Cla I site:

\[ 5'\text{-A}\underbrace{\text{T}\downarrow\text{CGA}}_{\text{Taq I site}}\text{T-}3'. \]

Both enzymes consequently cut the same bonds at this site.

6.8

The alternative \(GAT\downarrow C\) site would produce AT cohesive ends. Their annealing would form two A–T pairs instead of two G–C pairs.

A–T pairs have two hydrogen bonds each, whereas G–C pairs have three. In the intended comparison, the AT junction is less stable and dissociates at a lower temperature.

The requested curve should therefore be described as a similar increasing, approximately sigmoidal curve shifted to the left:

  • At a given temperature in the transition region, the percentage of linear DNA would be higher.
  • The temperature at which 50% is linear would be lower.

The supplied information does not determine the numerical shift.

6.9

Let \(P\) denote the open phage DNA, \(F\) the open fragment, and \(R\) the intended circular recombinant containing one of each:

\[ \ce{P_{linear} + F_{linear} <=> R_{circular,annealed}}. \]

Each molecule supplies two compatible cohesive ends. Formation of the recombinant circle produces two cohesive junctions, each containing two G–C pairs: four newly formed G–C pairs in total.

Base pairing and stacking release heat, so

\[ \boxed{\Delta H<0}. \]

There is no release of water in this annealing reaction. The backbone nicks remain unless a subsequent covalent ligation step is performed.

6.10

The intended qualitative combination is

\[ \boxed{\text{low temperature; low DNA concentration; high ionic strength}.} \]

  • Low temperature: Cohesive-end association is exothermic, so cooling favours the associated recombinant.
  • Low DNA concentration: After one phage end joins one fragment end, the remaining two ends are tethered within the same combined molecule and can close it into a circle. At high concentration, they can instead encounter additional DNA molecules and form longer chains. Dilution suppresses these additional intermolecular encounters relative to intramolecular closure.
  • High ionic strength: Salt, such as sodium chloride ([Na+].[Cl-]), screens electrostatic repulsion between the negatively charged phosphate backbones and stabilizes the paired cohesive ends.

The concentration argument must therefore include both ends of each molecule. Applying mass action only to the first bimolecular association would overlook chain formation.

“Low concentration” does not mean unlimited dilution: excessive dilution suppresses the initial phage–fragment encounter and favours separate self-circularization. The useful condition is sufficiently dilute to suppress multimer formation while still permitting the initial association.

The solution was checked against the official solution.