Quanave Cenral Dogma I Reference hp//book.bonumbers.org Inaon ranscrpon RNA polymerase and ranscrpon Facor (F) s bnds o promoer regon of DNA ranscrpon Meenger RNA, mrna, s produced and ranspored o Rbosomes ranslaon mrna read by Rbosomes o produce s Basc Sze and Geomery hp//book.bonumbers.org/wha-s-he-range-of-cell-szes-and-shapes/ Volume one lre = 1L = 10 3 m 3 ; E.Col sze s abou 1µm, and he volume s esmaed o be V Ecol! 1µm 3 = 10 18 m 3 (1µm 3 10 6 m µm 1 ( ) 3 = 10 18 m 3 ). Concenraon 1M = 1mol L 1, bu 1mol = 6.023 10 23, so 1M = 6.023 10 26 m 3, In agnmen 4, you wll show ha 1nM one molecule n a volume of 1µm 3. MaOne Dalon = Ma of Hydrogen = 1Da = 1.66 10 27 kg = 1.66 10 24 g, and one mole (Avogadro s number N A = 6.023 10 23 ) of hydrogen weghs 1 gram = 1g (( 1Da) 1 = ( 1.66 10 24 g) 1 = 6.022 10 23 g 1 ) 1g = 6.023 10 23 Da. One amno acd (aa) s 100Da = 1.66 10 25 kg 1aa 100Da. An average (gene) has abou 300 amno acds 1 300 aa. One base par (bp) s 650Da 1 bp 650 Da Proen Concenraon hp//book.bonumbers.org/how-many-s-are-n-a-cell/ Daa shows ha he ma per un volume s abou 0.2 g ml, bu µm 1mL = 10 6 m 3 = 10 6 m 3 10 6 m s n cell s M = 0.20 g Da ml 6.023 1023 3 = 10 12 µm 3, so Average Ma concenraon of g 10 12 ml µm = Da 3 1.2 1011 µm 3 Average Number concenraon of s n cell s found by N = M 1 100Daaa 1 300aa = 4 1 106. Noe ha he answer n µm 3 he webpage s wrong! here wll be a smlar queson on he agnmen.
mrna n Cells hp//book.bonumbers.org/how-many-mrnas-are-n-a-cell/ Expermenally, s known ha a bacerum has abou N bacera mrna 10 3 mrna, and a mammalan cell has abou N mamalan mrna 10 5. I s known ha he number of mrna n a cell remans relavely consan. I seems ha mrna concenraon scales wh sze. s Bacera Le s aume ha he concenraon n a cell s 4 10 6, µm 3 whch means ha a bacerum of volume 1µm 3 has abou 4 10 6 s. I s known ha, on average, bacera dvde every hour. hs means ha he number of s mus double every hour, so a cell mus produce 4 10 6 new s every hour, gvng a rae dn bacera = 4 106 s s 10 3. Aume now ha one 3600s s mrna can be ranslaed o abou 1 per second, or r mrna 1 s mrna s, whch gves number of mrna N bacera mrna ( dn bacera / )/( r mrna s ) = 10 3. Mammalan Cell Aume a volume of 3000µm 3 or 1.2 10 10 s. Mammalan cells dvdes every 24 hr, dn mammalan = 1.2 1010 s 24hr 3600s hr 1.4 s 1 105, and s wh r mrna 1 s mrna s, N mammalan dn mrna ( mammalan / )/( r mrna s ) = 1.4 10 5. Proen and mrna degradaon hp//book.bonumbers.org/how-fas-do-rnas-and-s-degrade/ Preamble As menoned mrna number remans, more or le, consan, and descrbed by he equaon, d k mrna, where s he number of mrna (n a cell) d s he rae of ncrease (or decrease) of mrna n uns of mrna s 1, r mrna s he rae of mrna producon, n mrna s 1, and k mrna s he rae of mrna n un of s 1. How are mrna degraded? In lvng cells, mrnas are degraded by enzymes (s), such as Rbonucleases (RNase). Hgh concenraon of hese enzymes ncrease he mrna rae, k mrna. he rae s he nverse of he half-lfe of mrna, k mrna = ln2/τ 1/2 mrna. For mrna n lvng cells τ 1/2 mrna = 3mn o 10Hr NOE DNA are degraded by s called deoxyrbonuclease (DNase)
Seady-Sae () Soluon Afer a suffcenly long perod, he sysem reaches a seady sae () of consan mrna number, = d = 0 0 k mrna /k mrna Exac Soluon for mrna he soluon of d = = k mrna s ( 1 exp( k mrna )). Afer a long perod,exp k mrna /k mrna. ( ) 0 Proens n lvng cells are produced by Rbosomes ha reads he mrna code. As such, he producon rae s proporonal o he number of mrna,. he rae of ncrease (or decrease) of s s dn = r k N, where N s he number of s, r s he rae of produced by ranslaon per mrna, and has un of s 1, k s he rae of. How are s degraded? In lvng cells, s are degraded by enzymes (s), such as proeases. Hgh concenraon of hese enzymes ncreases he rae, k. he rae s he nverse of he of s, k = ln2/τ 1/2. For mrna n lvng cells τ 1/2 = 30mn o days. Seady-Sae () Soluon Afer a suffcenly long perod, he sysem reaches a seady sae () of consan number, N = N dn = 0 0 = r k N N = r /k Exac Soluon for mrna he soluon of dn N = N N = N = r k N s ( 1 exp( k )). Afer a long perod, exp k = r /k. ( ) 0 WHA IS HE PROEIN O mrna RAIO? hp//book.bonumbers.org/how-many-s-are-made-per-mrna-molecule/ From he las secon we showed ha a seady sae N = r /k, ans he rao of Proens o mrna n a cell s N / = r /k. For E. Col r s he ranslaon rae per mrna whch can be argued o be r 1s 1, and
k = 1/τ, where he average me s τ 1000s, whch gves N / 1000, and ndeed, expermens usually observe ha he number of s s abou 1000 mes of he number of mrna. HOW FAS DO PROEASOMES DEGRADE PROEINS? hp//book.bonumbers.org/how-fas-do-proeasomes-degrade-s/ Proeasomes (or proeases) comprsed abou 1% of oal bulk s. One proeasome degrades 0.05 5 s/mn, whch s equvalen o abou 1aa s 1. However, hs depends on concenraon and emperaure In-Vro Proens In vro (es ube) soluons usually conan proease conamnans ha wll degrade s. Bu unlke n lvng cells, here are no rbosomes o produce new s o replace he degraded s. he rae equaon becomes dn = k N, wh soluon N exp k ( ), where N 0 s he number of s a = 0. We can calculae he half-lfe, τ 1/2, of s, whch s he me akes for he populaon o decrease by 1/2, N /2= exp k ( τ 1/2 ) τ 1/2 = ln2 = 0.693 k k Example 1 For a ypcal n vvo sysem, he concenraon s 4 10 6 µm 3, and he half-lfe s abou τ 1/2 = 40mn, or k me akes for he concenraon o decrease o 10% s found by ( ), whch gves N = 0.1N 0 exp k = ln( 0.1)/k proen = ln( 0.1)/0.0173mn 1 = 133mn. = 0.693 τ, k = 0.0173mn 1. he 1/2 he above calculaon s done for a ypcal n vvo concenraon of 4 mm, a room emperaure = 300K, and proease concenraon of abou 1% of s, or 4 10 2 mm = 40µM. o esmae he rae a dfferen emperaures and concenraons, we use Arrhenus Equaon Arrhenus equaon for reacon rae reacon rae k A = A c,d ( ), where = Aexp ΔE /k B ( ) depends on he collson frequency, whch n urns depend on he concenraon, c and he dffuson coeffcen, D. In cla, s shown
ha we can expreed A = A 0, where A s a consan, s he emperaure, l s 0 l 2 he mean, or average, separaon dsance beween 2 adjacen s. In he equaon, ΔE s he reacon free-energy barrer, whch we wll make an educaed gue equals o abou he energy released by a sngle AP hydrolyss, ΔE = 4.8 10 20 J Dverson on AP hydrolyss AP + H 2 O ADP + phosphae + energy, energy released s ΔH AP = 29 kj mol 29 kj mol 1000J /kj 6.023 10 23 AP mol J 1 4.8 10 20 AP Example 2 Suppose he emperaure s lowered from room emperaure, = 300 K (27 C), o jus above freezng F = 277 K (4 C). We can calculae he low emperaure (Fnal, F) rae k,f from he nal () room emperaure rae, k, k, k,f k, = 0.0173mn 1, by wrng = A 0 l 2 = F ΔE exp k B andk,f = A 0 F l 2 ΔE exp k B, and akng he rao F exp ΔE k B F exp ΔE = F exp ΔE 1 1 k B F, we have aumed ha l k B remans he same. hs gves k,f k,f k,f = k F, exp ΔE 1 1 k B F 277K = 0.0173mn 1 300K exp 4.8 10 20 J 1 1.381 10 23 J K 1 277K 1 300K = 6.1 10 3 mn 1 Le s calculae he me akes an n vro of he same concenraon as example 1 o o 10% s orgnal value, N = 0.1N 0 ( ), whch gves N = 0.1N 0 exp k,f = ln( 0.1)/k proenf = ln( 0.1)/6.1 10 3 mn 1 = 377mn, whch s longer, bu no subsanally longer for a bochems ryng o keep a soluon sable. Example 3 Suppose he emperaure s kep a room emperaure, bu he were reduced by 100 mes. Aume ha he orgnal concenraon s
N N l = = 4 10 6 s µm 3, as shown n cla he average separaon s s = 4 10 6 µm 3 N s 100 mes more dlue, F = 1 N 100 average separaon of l F = k, = A 0 ΔE exp 2 l k B andk,f k,f k, k,f ( ) 2 ( ) 2 = l 2 l = 6.3 10 3 µm 2 F 2.9 10 2 µm = 0.0173mn 1 6.3 10 3 µm = 6.3 10 3 µm. he fnal dlued concenraon N V = A 0 l F 2 = 4 10 4 s µm 3. hs gves a fnal s = 4 10 4 µm 3 = 2.9 10 2 µm. ΔE exp k B, and akng he rao ( ) 2 ( 2.9 10 2 µm) = 2 8.16 10 4 mn 1. Le s calculae he me akes an n vro of he same concenraon as example 1 o o 10% s orgnal value, N = 0.1N 0 ( ), whch gves N = 0.1N 0 exp k,f = ln( 0.1)/k proenf = ln( 0.1)/8.16 10 4 mn 1 = 2820mn, or abou 47 hr. How many rbosomes n a cell Read he followng hp//book.bonumbers.org/how-many-rbosomes-are-n-a-cell/