More Protein Synthesis and a Model for Protein Transcription Error Rates

Transcription

1 More Protein Synthesis and a Model for Protein James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University October 3, 2013

2 Outline 1 Signal Patterns Example One Example Two 2

3 Abstract This lecture is going to talk more about protein synthesis and a model of transcription error rates.

4 Signal Patterns Example One Now let s look at some problems where the signal to turn on and off protein production follows a given schedule. Consider the model P =.04P + 8 P(0) = 0 Let s draw a careful graph of the solution to this model using response time as the time unit. The production schedule is Production is on for 2 response times Production is off for 4 response times Production is on for 2 response time Production is off from that point on.

5 Signal Patterns Example One SS is 8.04 = 200. When the signal is off the model reduces to simple exponential decay with half life t 1/2 = ln(2)/.04 and when the signal is on the model grows using the response time t r = ln(2)/.04. So the same time scale works.

6 Signal Patterns Example Two Consider the model P =.05P + 11 P(0) = 100 Production is on for 2 response times Production is off for 3 response times Production is on for 2 response time Production is off from that point on.

7 Signal Patterns Example Two SS is = 220. When the signal is off the model reduces to simple exponential decay with half life t 1/2 = ln(2)/.05 and when the signal is on the model grows using the response time t r = ln(2)/.05. So the same time scale works.

8 Signal Patterns Example Two Homework 44 Draw a careful graph of the solution to the models below using response time as you time unit. Label important points and lines The model is P =.6P + 300; P(0) = 10. Production is on for 2 response times Production is off for 1 response times Production is on for 3 response time Production is off from that point on The model is Q =.8Q + 80; Q(0) = 20. Production is on for 1 response times Production is off for 3 response times Production is on for 3 response time Production is off from that point on.

9 Let s go back to the underlying biology again. We are going to explain why the error rate in transcription has a frequency of about To do this, we are going to look carefully at how the proteins are assembled in the ribosome and use a lot of reasoning with rates of interactions. We don t really use calculus and first order differential equations at all, although the idea of rates is indeed closely related. So grab a cup of coffee or tea and settle back for the story.

10 Let s go back to the underlying biology again. We are going to explain why the error rate in transcription has a frequency of about To do this, we are going to look carefully at how the proteins are assembled in the ribosome and use a lot of reasoning with rates of interactions. We don t really use calculus and first order differential equations at all, although the idea of rates is indeed closely related. So grab a cup of coffee or tea and settle back for the story. Recall the nucleotides in the gene Y are read three at a time to create the amino acids which form the protein Y corresponding to the gene. RNA polymerase, RNAp, binds to the promoter region and messenger RNA, mrna, is synthesized that corresponds to the specific nucleotide triplets in the gene Y. Once the mrna is formed, the protein Y is then made.

11 Now each mrna carries triplets of nucleotides which correspond to amino acids. Recall there are 64 possible triplets and 20 amino acids. A transfer RNA, trna, protein carries an amino acid which corresponds to a triplet on the mrna. We show this below. Figure: MRNA and trna in more Detail

12 When the combined structure, trna plus amino acid, binds to mrna, it is inserted into the ribosome. The amino acid it carries is added to the chain of amino acids already obtained and the trna is ejected. The next (trna plus amino acid) complex bound to the mrna is then read and the process repeats. This is drawn in below:

13 There are other (trna plus amino acid) complexes in the soup of components, proteins and other things that is what the inside of the cell looks like.

14 There are other (trna plus amino acid) complexes in the soup of components, proteins and other things that is what the inside of the cell looks like. The closest incorrect (trna plus amino acid) complex breaks apart or disassociates at the rate k d. The rate that the correct (trna plus amino acid) complex is k c and since we know the correct one should be read instead of the incorrect one, we must have k d > k c.

15 There are other (trna plus amino acid) complexes in the soup of components, proteins and other things that is what the inside of the cell looks like. The closest incorrect (trna plus amino acid) complex breaks apart or disassociates at the rate k d. The rate that the correct (trna plus amino acid) complex is k c and since we know the correct one should be read instead of the incorrect one, we must have k d > k c. We know that errors in reading are nevertheless made and experimentally we know it occurs at the frequency of Let s see if we can explain this.

16 Let the correct trna complex be denoted by c and the correct mrna triplet it corresponds to be C.

17 Let the correct trna complex be denoted by c and the correct mrna triplet it corresponds to be C. The two bind to to create another complex we will denote by cc.

18 Let the correct trna complex be denoted by c and the correct mrna triplet it corresponds to be C. The two bind to to create another complex we will denote by cc. Once the trna is created, there is a probability p per unit time the amino acid attached to the trna will be linked to the growing amino acid chain.

19 Let the correct trna complex be denoted by c and the correct mrna triplet it corresponds to be C. The two bind to to create another complex we will denote by cc. Once the trna is created, there is a probability p per unit time the amino acid attached to the trna will be linked to the growing amino acid chain. If this happens, the freed trna unbinds from the chain and the C shifts to the next (trna plus amino acid) complex.

20 We can analyze this with a bit of mathematical modeling like this. trna + mrna k c bound complex p [c] + [C] [cc] correct amino acid. where k c is the rate at which c and C combine to form the complex cc.

21 We can analyze this with a bit of mathematical modeling like this. trna + mrna k c bound complex p [c] + [C] [cc] correct amino acid. where k c is the rate at which c and C combine to form the complex cc. The complex also breaks apart at the rate k c which we denote in equation form as k c [cc] [c] + [C].

22 Combining, we have the model k c [c] + [C] [cc] correct amino acid k c p

23 Combining, we have the model k c [c] + [C] [cc] correct amino acid k c p At equilibrium, the rate at which cc forms must equal the rate at which cc breaks apart.

24 Combining, we have the model k c [c] + [C] [cc] correct amino acid k c p At equilibrium, the rate at which cc forms must equal the rate at which cc breaks apart. The concentration of cc is written as [cc]. The concentrations of c and C are [c] and [C]. The amount of cc depends on how much of the needed recipe ingredients are available. Hence the amount made is k c [c] [C].

25 The amount of c and C made because cc breaks apart depends on how much cc is available. So this amount must be k c [cc].

26 The amount of c and C made because cc breaks apart depends on how much cc is available. So this amount must be k c [cc]. So at equilibrium, because the formation rate and disassociation rate are the same that we must have a balance k c [cc] = k c [c] [C]

27 The amount of c and C made because cc breaks apart depends on how much cc is available. So this amount must be k c [cc]. So at equilibrium, because the formation rate and disassociation rate are the same that we must have a balance k c [cc] = k c [c] [C] Solving we find the relationship [cc] = k c k c [c] [C] which is a really common equation to come up with in this kind of analysis.

28 We call the fraction k c k c the disassociation constant K c and so we can write [cc] = 1 K c [c] [C] at equilibrium.

29 We call the fraction k c k c the disassociation constant K c and so we can write [cc] = 1 K c [c] [C] at equilibrium. The incorporation rate of the correct amino acid is then R correct = ( concentration of bound complex cc) ( probability amino acid is linked to protein chain) = [cc] p = p K c [c] [C].

30 We call the fraction k c k c the disassociation constant K c and so we can write [cc] = 1 K c [c] [C] at equilibrium. The incorporation rate of the correct amino acid is then R correct = ( concentration of bound complex cc) ( probability amino acid is linked to protein chain) = [cc] p = p K c [c] [C]. The incorrect trna will be represent by d and it binds in a similar way to form the complex dc. The rates of combination and breaking apart are now given by k d and k d. The same reasoning as before gives us the model k d [d] + [C] [dc] incorrect amino acid k d p

31 And at equilibrium (we use the same reasoning!), we find [dc] = 1 K d [d] [C] where K d = k d k d. The rate of incorrect linking is then R incorrect = = incorrect rate correct rate p [d] [C] p [c] [C] /. K d K c

32 And at equilibrium (we use the same reasoning!), we find [dc] = 1 K d [d] [C] where K d = k d k d. The rate of incorrect linking is then R incorrect = = incorrect rate correct rate p [d] [C] p [c] [C] /. K d K c We can cancel the common p and [C] to get R incorrect = incorrect rate correct rate = [d] / [c] K d = K c K d K c [d] [c].

33 Now in a cell, we know from experimental data that the concentrations of the incorrect and correct trna s are about the same. Hence, [d] [c] 1 and we can say with a little bit of algebra that R incorrect K c K d = k c k d. k c k d

34 Now in a cell, we know from experimental data that the concentrations of the incorrect and correct trna s are about the same. Hence, [d] [c] 1 and we can say with a little bit of algebra that R incorrect K c K d = k c k d. k c k d From experimental data, we also know the on rate for binding of both d and c are limited by how molecules diffuse through the cell and because of that are about the same. Hence, k d k c 1 which leads to our final result. R incorrect k c k d.

35 Now recall these terms k d and k c are rates of breaking apart the complexes. So since the incorrect binding is weaker, we know k d > k c and so we know R incorrect < 1. We also can measure these rates and they have been determined to give the ratio k c 10 k 2. So the error rate, determined by an d equilibrium analysis is 100 times higher than the true rate 10 4! What is wrong with our analysis? How do we explain this?

36 Now recall these terms k d and k c are rates of breaking apart the complexes. So since the incorrect binding is weaker, we know k d > k c and so we know R incorrect < 1. We also can measure these rates and they have been determined to give the ratio k c 10 k 2. So the error rate, determined by an d equilibrium analysis is 100 times higher than the true rate 10 4! What is wrong with our analysis? How do we explain this? The way to handle this discrepancy is to introduce the idea of kinetic proofreading. And that is what the next lecture will be about!