1. Basics of hybridization

2. What is hybridization? n Complementary base pairing of two single strands of nucleic acid  double strand product u DNA/DNA u RNA/RNA u DNA/RNA

3. What holds the two strands together? n Hydrogen bonds between the base pairs n Hydrophobic interactions of stacked bases n van der Waals forces between stacked bases

4. Factors affecting the strength of strand pairing n Number of GC pairs vs. AT pairs n Mismatch n Length of hybridizing strands n [Salt] of hybridization solution n Temperature n Concentrations of denaturants

5. Factors affecting the strength of strand pairing n Number of GC pairs vs. number of AT pairs u The more H-bonds between strands, the more strongly they are held together F 3 H-bonds between G and C F 2 H-bonds between A and T u So…the more GC pairs, the more H-bonds between strands

6. Factors affecting the strength of strand pairing n % Mismatch u the greater the lack of complementarity, F the fewer hydrogen bonds u the lower the strength of the hybrid

7. Factors affecting the strength of strand pairing n Length of hybridizing strands u the longer the strands, F the more hydrogen bonds and F the more hydrophobic interactions, so u the greater the strength of the hybrid

8. Factors affecting the strength of strand pairing n [salt] of solution n  [salt]   strength of the hybrid u negative charges of the phosphate moieties of the sugar-phosphate backbones repel each other u + ions from salts in solution act as counterions to reduce repulsion F Monovalent cations (Na + ) F Divalent cations (1 mM Mg ++ = 100 mM Na+) –Why does [Mg ++ ] affect specificity of PCR priming?

9. Factors affecting the strength of strand pairing n Temperature u heat increases the kinetic energy of each of the two strands u sufficient heat makes kinetic energy > H-bond energy u strands separate

10. Factors affecting the strength of strand pairing n pH u [OH - ], ~pH 12 F enolic hydroxyl groups on bases ionize keto-amino H-bonds disrupted Keto Enol ionizes GG G

11. Factors affecting the strength of strand pairing n pH u [OH - ], ~pH 12 F N3 on thymine and N1 on guanine lose their hydrogens and resulting negative charge is delocalized over the ring

12. Factors affecting the strength of strand pairing n pH u [OH - ], ~pH 12 F enolic hydroxyl groups on bases ionize keto-amino H-bonds disrupted n Concentration of denaturants u formamide, urea

13. Factors affecting the strength of strand pairing n And in our case... u The presence of the alkaline phosphatase enzyme causes some steric hindrance.

14. Combined effects of these factors can be expressed as equations for the Tm – points to be covered n What is Tm? n Equation to estimate Tm for DNA oligonucleotides n Equation to estimate Tm for polynucleotides

15. What is Tm? n Tm = temperature of melting or separation of strands u Tm is a function of the DNA fragment or RNA strand under consideration and the solution in which the hybridization is occurring. F Changing the temperature does not change the Tm!

16. What is Tm? n For complementary oligonucleotides (10 - 23 nt) u Temp at which 50% of complementary molecules exist as single strands 50% 5’ - - - - - - - - - - - - - 3’ 3’ - - - - - - - - - - - - - 5’ 50% 5’ - - - - - - - - - - - - - 3’ 3’ - - - - - - - - - - - - - 5’

17. What is Tm? n For complementary polynucleotides (>~25nt) u Tm is the temp at which 50% of hydrogen bonds within any one hybrid are broken

18. Combined effects of factors contributing to strength of a hybrid can be expressed as equations for Tm n for DNA oligonucleotides in 1.0M Na + Tm ( o C) = 4 (G+C) + 2 (A+T) n Note: how does this equation account for u Length? u Difference in strength between G/C vs. A/T bonds? u The conditions of the solution?

19. Combined effects can be expressed as equations for Tm n for DNA polynucleotides and oligos as short as 14 nt Tm = 81.4 + 16.6 log [(M + )/1+0.7(M + )] + 0.41 (%G+C) - 600/L - %mismatch - 0.65 (% formamide) M + = monovalent cation concentration L = length of probe sequence

20. Tm for polynucleotides (cont’d) n How does the equation on the previous slide account for u Length? u Difference in strength between G/C vs. A/T bonds? u The conditions of the solution?

21. Membrane hybridization n One nucleic acid component is affixed to membrane; the other is in solution u 14;18 translocation: samples affixed; probe(s) in solution n Membrane material binds DNA or RNA u nylon u charged nylon u nitrocellulose

22. Typical steps in membrane hybridization n blocking or prehybridization n hybridization n wash or rinse n Visualization

23. Blocking/prehybridization n Why? u Remember, membrane binds nucleic acid, so F labeled nucleic acid in hybridization solution can bind everywhere on membrane   background

24. Blocking/prehybridization n How? u Membrane with affixed nucleic acid is bathed in blocking solution at hybridization temperature u Components of blocking solution bind non-specifically to membrane to prevent labeled nucleic acid from binding except to complementary strands

25. Blocking/prehybridization n common blocking agents u sodium dodecyl sulfate (SDS) u nonfat dry milk u bovine serum albumin u Ficoll (carbohydrate polymer) u polyvinylpyrollidone (PVP)

26. Hybridization n What? u Labeled nucleic acid in solution is allowed to anneal to affixed complementary strands n Conditions u Must be determined empirically u Hybridization solution includes F [Salt] determined from Tm formulas F Membrane blocking agents F Labeled nucleic acid If necessary, denatured by –High temperature (95 o C) or –Alkaline (high pH) conditions

27. Hybridization n Conditions (cont’d) u Temp set below Tm to optimize rate of hybridization F oligonucleotides: 15 o below Tm F polynucleotides: 15-35 o below Tm

28. Wash/rinse n Why? u To remove labeled probe/sample that is F in excess F non-specifically bound F bound with loose complementarity

29. Wash/rinse How? n Bathe membrane in solution lacking labeled probe/sample n Use stringency conditions that minimize non-specific hybridization u stringency = likelihood that two strands will separate

30. Wash/rinse How? n Be aware that wash conditions for oligonucleotide and polynucleotide hybridizations differ because: u oligonucleotide hybrids are not in equilibrium F Once separation occurs, if hybridizing strand is not present in excess, hybridization is unlikely to recur. A dilution effect is sufficient. u polynucleotide hybrids are in equilibrium within the strand F a temperature, low salt, or denaturant effect is necessary

31. Choosing wash conditions (cont’d) n To wash oligonucleotide hybridizations u Use stringency similar to or lower than hybridization condtions F Same or lower temperature F Same or higher salt concentrations u Short time periods

32. Choosing wash conditions u To wash polynucleotide hybridizations (equilibrium) F raise stringency conditions to make it harder for imperfect hybrids to remain annealed F perform washes just below the Tm u  stringency   likelihood that two strands will separate F Lower the salt concentration F Raise the temperature F Include denaturants

33. Visualization n requires a visible signal u radioactive u non-radioactive, enzyme linked u non-radioactive, non-enzymatic F e.g., use of fluorescent label n for enzyme-linked signal generation u additional block and rinse steps required F avoid conditions that will disrupt hybrids

34. Reminder n Combine your knowledge of hybridiation with your knowledge of Southern transfer.