Unformatted text preview:

Lecture #3: Nutrient Assimilation 1, Ch.6 and box 37.3 Hb – O2 Binding Curveo What’s important about this curve? Sigmoid: due to cooperative binding, molecules with subunits. Why? Has to do with 2 diff. parts of the curve.o 1.Steep part says that as your cells use more oxygen you get a must larger increase in the amount of oxygen unloaded by the hemoglobin.o 2.Top part of curve says when you have a wide range mostly the hg is saturated. o For a organism with less oxygen curve would lie to the lefto For an organism that has/needs A LOT of oxygen i.e. a bird the curve would lie tothe right (b/c you want to unload the hemoglobin more) Review of O2 carrying molecules, e.g., Hbo Subunits = cooperative bindingo Cooperativity = sigmoid binding curveo Two critical sections of curveo Conditions which demand increased O2 result in shift of curve to right Decreased affinity increased P50o Several O2 binding proteins in animals All exhibit cooperative binding and different P50 values to match environment P of O2 Unload very quickly Enzyme carbonic anhydrous * must know this reactiono Proton concentration we measure as the pHo When you hyperventilate you are eliminating more carbon dioxide – organism body fluid pH will increase. This is why a hyperventilating person breathes into a bag  Nutrient Assimilation Single celled organismso Must transport molecules from the environmento Simple molecules through cell membrane via transport or diffusiono Complex molecules engulfed by other mechanisms Multicellular organisms: Two considerationsHb saturation at wide range of P O2o 1. Gathering and ingesting the nutrientso Breaking down complex molecules Extracelllar digestiono 2. Transporting simple molecules into the cells of the organism Examine how solute molecule X enters a cello We must be able to measure [X] Experiment: Begin with cells in a medium suitable for survival and growth. Now add the solute of interest X and know the final [X] in the medium. Let cells incubate with X for various amounts of time and then measure the amount of X inside the cells (or measure amount still in the medium)! Then plot the results in a graph – time vs. Amount of X in cells. o You will see that X eventually reaches a max value (in cells). Why is this? At some point the amount of X inside the cell will equal the concentration outside thecell and if simple diffusion is the mechanism going on, it will stop.o You cannot tell whether the X is entering the cell through active transport or diffusion. Point of this experiment is to determine how the X is entering the cell. Why would X in cells reach some constant amount?o Equal concentrations in and out Probably by diffusiono Cells have some feedback mechanism to stop transport Probably by some transport mechanismo PROBLEM: difficult to measure intracellular concentrations of solutes! Let’s now plot the results, one plot for each initial [X]o Make three different tubes with high/medium/and low concentrations of X. Three differenttest tubes, three different experiments. Difference: three different concentrations of X. o When you increase the amount of concentrations the rate of entry is increasing. o The measure of the rate of increase is the slope (incline in the beginning especially). TIME (minutes)AmountOfXInCells(moles)High [X]o Ex: these lines are calculating the rate. o Direct proportion between concentration of X and rate of entry – this is the EXACT piece of data we need to figure out how X enters the cello Important: you must be able to describe this experiment, say how the graph is set up etc. This graph doesn’t tell us anything the rates do Rate of entry of X into the cells. Slope – umoles/min One possible result: Rate as function of [X]mediumo Is a graph with directly proportional initial rate of uptake vs, [X] initial o Has no maximum rate: no matter the concentration the rate of X will go up Another possible result: Rate as function of [X]mediumo Some solutes will eventually have a max. rate of entry (notice how slope levels off) To interpret this data, let’s think about the Chesapeake Bay Bridge on a Friday in the summer!o Ocean city on a Friday: say there is 5 toll boothso Each tollbooth can do 5 cars per minute. o Ex: at 1:04 25 cars per minute get to the bridge how many get across -25 cars get by per minuteo At 1:20 45 cars get to the bridge 25 cars get by per minute Rate of cars going across rate levels off at 25 cars o There is a max rateo Would there be a max rate if there was no toll booths? Noo As you can see these examples are picture examples of diffusion (no toll booths)o Toll booths, molecules that need transport proteins and membranes to get across  If you see the graph above it symbolizes no transport sites, simple diffusion through the membrane. Small nonpolar molecule – does not need a transport site. If you see the graph above it symbolizes specific transport sites limiting the maximum rate.Polar molecule that will not dissolve through the membrane it must evolve transport sites.  Must be nonpolar to dissolve through membrane Oil/Water Partition Coefficiento Add solute X to a water and oil container (oil and water are separated from each other). A nonpolar molecule will go into the oil. Polar molecule will go into the water. Partition means it divides –some.  O/W PC = [S]oil [S]water Simple Diffusiono Nonpolar molecules would have high p.c. …polar molecules would have low p.c. o **Understand how this graph was generated.  If two molecules have the same p.c. the smaller molecules will diffuse faster. Look at examples of this above. Some things have the same p.c. but some diffuse faster.  Problem: the H2O at the top of this chart. Water does not just diffuse through the membrane Differential permeability - the relative abilities of different molecules to diffuse across membranesRate ofDiffusionnmoles/min/ 106 cellsLog solubility in lipidH2OOil/Water Partition Coefficiento Concepts:o Diffusion of nonploar solutes and gases occurs by simple diffusion.o No specific transport siteso Depends upon the C (difference in concentration) and/or DE of the solute (electrochemical gradient)o Goes in both directions across the membrane 1.Simple diffusion – molecule n.p. 2.Rate of diffusion depends on the concentration gradient as well as the oil/water p.c. Goes in both


View Full Document

UMD BSCI 207 - Lecture #3

Documents in this Course
Notes

Notes

15 pages

Neurons

Neurons

27 pages

Exam 3

Exam 3

5 pages

Motility

Motility

19 pages

Final

Final

20 pages

Exam 3

Exam 3

4 pages

EXAM 2

EXAM 2

12 pages

DNA

DNA

11 pages

Load more
Download Lecture #3
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Lecture #3 and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Lecture #3 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?