It is possible to measure the rate of the reaction by following the consumption of the substrate, as you suggested, but there is a drawback to doing that. For steady-state kinetic measurements, you must measure the initial rate. This usually means that no more than 5-10% of the substrate can be consumed, particularly when the substrate concentration is low relative to the Km, because the reaction slows down significantly as the substrate is consumed. It is difficult to measure such small changes in substrate concentration accurately and precisely.

A better method is to measure the formation of the product, because the reaction starts with no product, so you can more easily measure the change in its concentration over time.

Measuring the kinetics could get a bit complicated in this case because the radioactivity of the bands decreases as their size decreases, if they are uniformly labeled along their length, so you have to correct for that when you calculate their concentrations. You can also avoid this issue by only measuring the reaction for as long as one nucleotide is removed. Another way to avoid this problem is to use DNA that is only labeled at the 5' end, either with radioactivity or a fluorescent dye, so that all products give the same signal no matter how long they are.

Also, the nature of the substrate changes, i.e. it gets shorter, as the nucleotides are removed from one end. The enzyme may not work at the same rate on a short substrate as on a long one. This problem can be avoided if you start with a long enough substrate and do not let the reaction proceed very far.

Collect evenly spaced time points and plot the product concentration with time for each substrate concentration. Measure the slope of this plot that goes through the origin and is tangent to the start of the curve. That is the initial rate.

It is possible to measure the rate of the reaction by following the consumption of the substrate, as you suggested, but there is a drawback to doing that. For steady-state kinetic measurements, you must measure the initial rate. This usually means that no more than 5-10% of the substrate can be consumed, particularly when the substrate concentration is low relative to the Km, because the reaction slows down significantly as the substrate is consumed. It is difficult to measure such small changes in substrate concentration accurately and precisely.

A better method is to measure the formation of the product, because the reaction starts with no product, so you can more easily measure the change in its concentration over time.

Measuring the kinetics could get a bit complicated in this case because the radioactivity of the bands decreases as their size decreases, if they are uniformly labeled along their length, so you have to correct for that when you calculate their concentrations. You can also avoid this issue by only measuring the reaction for as long as one nucleotide is removed. Another way to avoid this problem is to use DNA that is only labeled at the 5' end, either with radioactivity or a fluorescent dye, so that all products give the same signal no matter how long they are.

Also, the nature of the substrate changes, i.e. it gets shorter, as the nucleotides are removed from one end. The enzyme may not work at the same rate on a short substrate as on a long one. This problem can be avoided if you start with a long enough substrate and do not let the reaction proceed very far.

Collect evenly spaced time points and plot the product concentration with time for each substrate concentration. Measure the slope of this plot that goes through the origin and is tangent to the start of the curve. That is the initial rate.

Adam B Shapiro Thank you so much for such a perfect explanation for my doubt. I always receive tremendous support from your suggestions on research gate. I will do it exactly what you have mentioned in your suggestions. Well, the oligos here is fluorescein labeled at 5' end, but here you can see that multiple products have been formed (in the same lane) after removal of each nucleotide by the enzyme. In such case, which product I should really look at ? Is it only the first product I should care about or should I measure all the products appeared here (formed) in the same lane ? It's bit confusing me here. I mean, is there any particular product you are talking about ? Please shed light on it.

The first band under the substrate band represents removal of one nucleotide presumably (call that band 1), and the band below that represents removal of 2 nucleotides (call that band 2). Since the bands should all have the same fluorescence intensity/mole of DNA or RNA (because the substrate is end-labeled at the 5' end), the product concentration is therefore directly proportional to the intensity of the band 1 + twice the intensity of band 2. I think you should count both bands, but remember that the amount of product should not exceed 10% of the amount of substrate if you are measuring the initial rate. Under those conditions, there may not be enough of band 2 to measure.

You will need a standard curve consisting of different amounts of substrate in order to convert band intensity to product concentration.

Sourav Roy

Prof. Shapiro, I tried with three concentration of the enzyme (0.5 nM, 5 nM and 10 nM) fixing the substrate concentration (50 nM) in order to determine the initial velocity. So my question here is how to decide what substrate concentration I have to consider to keep it constant. here I have used 50 nM oligo substrate to start the experiment. How one can decide the concentration of the substrate when our enzyme concentration is the variable to determine the initial velocity. In this experiment, I could see that 5 and 10 nM enzyme is rapidly consuming the substrate. While, 0.5 nM enz showed slowed degradation. Image of the result is attached. Looking forward to hearing from you. Adam B Shapiro

You just have to experiment to find the right ranges of substrate and enzyme concentrations. It is important for steady-state kinetics that the enzyme concentration be much lower than the substrate concentration, so that the free substrate concentration is not significantly depleted by binding to the enzyme. 50 nM substrate with 0.5 nM enzyme is a good combination in this respect.

You also to have consider the time scale of the reaction to be able to measure the initial rate. You may run into a problem of insufficient sensitivity for product detection, since the product band is barely detectable when starting with 50 nM substrate. You will have to detect single-digit nM product. Increasing the substrate concentration will help with this problem, but you also have to consider what the substrate Km is. To measure the Km properly, you need to use substrate concentrations both above and below it. If the Km is too low, you may not be able to measure it by this method.

Adam B Shapiro Thank you your reply ! I was worried exactly about the same issue what you have mentioned lastly, that if the Km of this enzyme is below 10 nm, 10 nM, 5nM or so, then this method of detection of the substrate is not appropriate to use. In that case what would you recommend us ! Should I move to radioactive labeled substrate as it may be more sensitive method than this very FAM labeled substrate ! Please shed some light on this issue. Thank you !

Before exploring more sensitive detection techniques, first determine whether they are needed. Try measuring the Km at higher DNA concentrations than you have been using so far.

If you can measure complete progress curves (product concentration versus reaction time) for multiple DNA concentrations at an enzyme concentration that is

Prof. Shapiro,

That's really a great suggestion !

Prem

Adam B Shapiro Dear Prof Shapiro,

I am calculating the activity of my 3' exonuclease enzyme. There is one paper that has discussed the calculation of a type of exonuclease activity (Ref: doi:10.1016/bs.mie.2019.05.004). The way they have calculated the activity, I am not sure if this is the unit of enzyme activity that can be used for calculation of kinetic parameters such Km and Kcat. I have attached a screenshot of the calculation they described in the paper. Please suggest me if it can be done through this way. Or any modification is required in this equation.

The factor 1.5 in the summation must be the source of the picomoles. Otherwise, the quantity is a unitless ratio, the sum of the ratios of each band intensity (other than the highest one, because it is the substrate) to the total band intensity in the lane. Band 2 has had one base removed, band 3 has had 2 bases removed, and so forth. Therefore, I assume that the reaction started with 1.5 picomoles of substrate.

This method allows you to calculate the rate of the enzyme reaction by allowing you to calculate the number of picomoles of substrate formed at any given time prior to quenching the reaction.

In order to calculate Km and kcat using the usual Michaelis plot procedure, you are going to have to vary the substrate concentration (which changes the 1.5 factor) and measure the amount of product released at various time points, while keeping the enzyme concentration constant. You will then measure the initial rate of reaction at each substrate concentration, prepare a Michaelis plot, and fit the data to the Michaelis-Menten equation to get Km and Vmax. Finally, you will divide Vmax (picomoles/second) by the enzyme concentration (picomoles) to get kcat (s-1).

This will be a lot of work using an endpoint assay and a gel electrophoresis readout.

There is an assumption being made in this analysis, which is that all the bands would have equal intensity if they were all present in the gel in the same amount. This is probably true if the 5' end of the substrate has the fluorescent label attached, but it is probably not true of you are staining the nucleic acid after running the gel, because smaller nucleic acid molecules bind less stain than larger ones.

Adam B Shapiro Thank you very much for your valuable and explanatory suggestions. You are absolutely right. In a 30 ul total reaction mixture, they have used 50 nM of FAM labelled substrate and so multiplying these (conc. and vol.) give rise to this factor 1.5 (i.e picomole of the substrate used in the total reaction). I have few concerns in the paragraphs you mentioned above.

1) As you said "This method allows you to calculate the rate of the enzyme reaction by allowing you to calculate the number of picomoles of substrate formed at any given time prior to quenching the reaction". In this, are you talking about product formed instead of "substrate formed"?

2) Secondly you mentioned " In order to calculate Km and kcat using the usual Michaelis plot procedure, you are going to have to vary the substrate concentration (which changes the 1.5 factor) and measure the amount of product released at various time points, while keeping the enzyme concentration constant. You will then measure the initial rate of reaction at each substrate concentration, prepare a Michaelis plot, and fit the data to the Michaelis-Menten equation to get Km and Vmax". So here, I understood that I need to vary the substrate conc. while keeping the enzyme conc. constant and that will change the factor as the substrate conc. is changing. Now, if suppose I already know that my substrate is changing into 10% of product within 1 min at the minimum detectable substrate conc. (say 50nM), then do I still need to carry out reactions at various time point (say 1 min 5 min 10 min etc) for each substrate concentrations ? or just same 1 min can be used for every reactions with varying substrate conc. ? I would then be plotting [Product] vs time (min) to get the initial rate and then this Initial rates at different time points will be plotted against the substrate conc. to determine Km and Vmax. Please correct me if I am wrong.

I understand that it will be lot of work, but I have no alternatives than this as of now.

Thank you again

Sincerely,

Prem

1. Yes, I meant product, not substrate.

2. This answer is in several parts.

a. It is not advisable to use a single time point and assume that it represents the initial rate, even if only 10% of the substrate is consumed, because the assumption may be wrong.

b. If the reaction is consuming 10% of the substrate in just 1 minute, it would be advisable to slow the reaction down by reducing the enzyme concentration, giving you time to measure multiple time points.

c. There is another important point about measuring the Michaelis-Menten constants that has not yet been addressed: the requirement that the substrate concentration be far in excess of the enzyme concentration. The reason for this requirement is so that the free substrate concentration at steady-state is essentially identical to the total substrate concentration. In other words, the fraction of the substrate bound to the enzyme is negligible. With only 50 nM substrate, the enzyme concentration will probably have to be

@Adam B Shapiro I am facing an issue while using the lower Substrate concentration. Suppose I will plot the initial rates v/s varying substrate concentrations. In this case I need to keep the enzyme concentration constant. For example if I use 200 nM substrate and I established that 2 nM enzyme works well with this concentration. Now what if I lower the substrate concentration say 20 nM, for this [S] the 2nM enzyme will lead to a very rapid degradation, if I keep the 2nM enzyme constant throughout the substrate concentration stretch. And my second question is to calculate Kcat we need Vmax/[Ei] so what is this Ei ? Is this the enzyme concentration used in the reaction mixture tube or something else ? Please suggest.

Adam B Shapiro I am facing an issue while using the lower substrate concentration. Suppose I will plot the initial rates v/s varying substrate concentrations. In this case I need to keep the enzyme concentration constant. For example, if I use 200 nM substrate and established that 2 nM enzyme works well with this concentration of substrate. Now, what if I lower the substrate concentration say 20 nM, so for this [S], the 2nM enzyme will lead to a very rapid degradation if I keep 2nM enzyme constant throughout the substrate concentration (20, 50, 100, 200 300 400 500 nM) stretch. And my second question is to calculate Kcat we need Vmax/[Ei]. So what is this Ei ? Is this the enzyme concentration used in the reaction mixture tube or something else ? Please suggest.

1. It is not necessary to keep the enzyme concentration the same for every substrate concentration. What is necessary is that the initial rate of the reaction is directly proportional to the enzyme concentration. That way, you can use different enzyme concentrations as needed for each substrate concentration, then divide the rate by the enzyme concentration to correct for the differences in enzyme concentration. You should do an experiment to establish that the initial rate is directly proportional to the enzyme concentration at one substrate concentration.

2. kcat = Vmax/[E], where [E] is the enzyme concentration. I'm not sure why there is an i there. If you use the method described in answer #1, whereby the measured rates have already been divided by the enzyme concentration, then kcat = Vmax instead.