“…When the hammerhead helices are short, product oligonucleotides rapidly dissociate from the ribozyme after cleavage and thus do not influence the cleavage rate+ However, when the helices are sufficiently long for the products to remain stably bound, the overall cleavage rate is affected+ When ribozyme is in excess, stably bound products only slightly increase the rate of observed cleavage and slightly decrease the extent of reaction due to the slow reverse reaction+ However, when substrate is in excess, stable products can dramatically influence the cleavage rate because their release becomes rate-limiting+ Because product RNA duplexes as short as five base pairs can have dissociation rates at 25 8C slower than 1 min Ϫ1 , many hammerheads will show slower cleavage rates when measured in substrate excess than when measured in ribozyme excess+ Indeed, there are numerous examples in the literature where the multiple-turnover cleavage rate at saturation was not assigned to a specific step of the reaction and the rate determined probably reflected product release (Koizumi et al+, 1989;Goodchild, 1992;Hendry et al+, 1992;Paolella et al+, 1992;Slim & Gait, 1992;Taylor et al+, 1992;Hendrix et al+, 1995;Holm et al+, 1995)+ A burst experiment is useful for determining whether the rate-limiting step for a hammerhead reaction is k 2 or one of the product release steps+ A burst reflects the fast appearance of products during the first turnover of the ribozyme that is followed by a slower appearance of products representing the subsequent rate-determining step of the pathway+ Experimentally, a burst is easiest to detect when the substrate concentration is sufficiently high to be saturating throughout the experiment and a 3-10-fold excess of substrate over the ribozyme is used+ Figure 7 gives two simulated data sets for a hammerhead that is limited by product release when a ratio of substrate to ribozyme of 10:1 was used+ If all of the ribozyme can participate in the reaction, the formation of 10% product is equivalent to one turnover+ In Figure 7A, several time points were taken in the first turnover, clearly revealing the burst of product formation+ The rates of each phase as determined from this data set are 1+1 min Ϫ1 for the burst reflecting k 2 followed by 0+67 min Ϫ1 for subsequent turnovers reflecting release of one of the products+ This burst behavior is more difficult to detect if insufficient data points are taken during the first turnover (Fig+ 7B)+ Its existence can only be surmised by showing that the steady-state rate does not extrapolate to the origin at zero time+ Similarly, if the substrate is in greater excess, not all the ribozyme is active, or if saturation is not maintained throughout the reaction, the burst phase may also be difficult to detect+ This example stresses the importance of carefully defining the first turnover especially when the rate of product release may be close to the rate of the chemical step+…”