1 / 20

Tissue Turnover and Growth

Tissue Turnover and Growth. Turnover assessed via source switching experiments. Fitting exponentials. Describe growth of critter as: k = ln( M t / M 0 )/ t where k is specific growth rate constant, t is time, M t is mass at time, M 0 is starting mass.

faulknert
Download Presentation

Tissue Turnover and Growth

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Tissue Turnover and Growth

  2. Turnover assessed via source switching experiments

  3. Fitting exponentials Describe growth of critter as: k = ln(Mt/M0)/t where k is specific growth rate constant, t is time, Mt is mass at time, M0 is starting mass. Model isotope incorporation during growth and turnover as: d/dt = -(k+m)(t-E) where m is metabolic (turnover) rate constant, d/dt is the change in isotope value with time, t is the isotope value at time t, and E is isotope value at equilibrium with new input. Integrating between t = 0 and t at equilibrium yields: t-E = (0-E )e-(k+m)t or t = E + (0-E )e-(k+m)t or t = E + (0-E )e-t where  is overall rate constant for isotopic incorporation Half-life (time for signal to shift by 50%): t1/2 = (ln2)/ 

  4. Dealing with Growth and Turnover (MacAvoy et al. 2005) 13 weeks old on day 1: sexually mature, adult?

  5. Dealing with Growth and Turnover (MacAvoy et al. 2005) Muscle Liver Blood Expected due to growth alone (m=0): t = E + (0-E )e-(k+0)t

  6. Dealing with Growth and Turnover (MacAvoy et al. 2005) Muscle Liver Blood Expected due to growth and turnover (m≠0): t = E + (0-E )e-(k+m)t

  7. Dealing with Growth and Turnover (MacAvoy et al. 2005) m = k + (ln[(t-E )/(0-E )])/t

  8. Dealing with Growth and Turnover (MacAvoy et al. 2006) Adults (?) given 120 to equilibrate to control diet before start of experiment. No change in size during experiment.

  9. Mouse Rat Dealing with Growth and Turnover (MacAvoy et al. 2006)

  10. Dealing with Growth and Turnover (MacAvoy et al. 2006)

  11. Reaction Progress Variable (Cerling et al. 2007)

  12. Reaction Progress Variable (Cerling et al. 2007) Fractional approach to equilibrium: scale 1 to 0 Linearizes and normalizes exchange: y = mx + b ln(1-F) = - t + b eb is the size of the fractional size of the pool (i.e., 1 when b=0)

  13. Reaction Progress Variable (Cerling et al. 2007) Can deal with up to three pools (j) of size (ƒ) with different s Each pool changes as:

  14. Reaction Progress Variable Pool 1 size: 0.7 t1/2: 2 days Pool 2 size: 0.3 t1/2: 20 days

  15. Reaction Progress Variable Example 1: Mouse breath (combining experiments)

  16. Reaction Progress Variable Example 2: Horse hair (multiple pools)

  17. Reaction Progress Variable Example 2: Horse hair (multiple pools)

  18. Reaction Progress Variable Example 3: Warbler blood (dealing with delays)

  19. Reaction Progress Variable Example 4: Horse diet from horse hair (inverse modeling) Each pool changes with time incrementally as:

  20. Reaction Progress Variable Example 4: Horse diet from horse hair (inverse modeling)

More Related