Flagella Lab
description
Transcript of Flagella Lab
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Lab 2
Flagella Regeneration
in Chlamydomonas reinhardtiior how simple microscopy can be used to calculate rates of assembly of macromolecules into cellular structures
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Andrew, Jiayun, Michael, Madeline
Fri Allens sec.
Bharat, John, Russell & Fahd
Fri Kais sec
Rubi, Michelle, David & Raj
Fri Kais sec
Chlamy fixed in Lugols Iodine
1000X phase contrast
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experiment
Remove flagella with a pH shock, pH 7.5 to 4.5 for 60s then back to pH 7.5In teams of 4, design an experiment to determine the effect of a drug on flagella regeneration.Two flasks, one control, one with a drug.Possible drugs are calcium, lithium, cycloheximide, actinomycin D or caffeine.Illuminate for one hour (or more if needed), taking and fixing samples every 15 min.Mount samples and measure flagella lengths.Determine rate of flagella growth in micrometers per minute. -
the flagella
Electron MicrographDiagramatic Drawing
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The axoneme is composed of a 9 + 2
structure of microtubules.
Each of the nine outer are partial doubles,
one complete microtubule with 13
protofilaments and one partial microtubule
with 10 protofilaments.
The two inner are complete microtubules.
From Iowa State University Agronomy 317
- Based on your rate of flagella regeneration, calculate the number of tubulin monomers assembled per minute.How would you make this calculation?
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Actinophrys,
a heliozoan protist.
The heliozoan has a spherical cell body with thick radiating strands of cytoplasm called axopodia, which are supported by microtubules. When smaller organisms bump into the axopodia, they stick to the axopodia and are then moved down the axopodia to the cell body, where they are phagocytosed.http://www.microscopy-uk.org.uk/mag/imgfeb02/acti3web.jpg
- Chlamydomonas adheres to the axopodia by its flagella but not
its cell body, and has evolved the ability to escape by shedding
its flagella. As a result, the heliozoan has flagella for dinner,
while the Chlamydomonas cell floats away and lives to grow new
flagella.
Ref: Pickett-Heaps, J. and Pickett-Heaps, J. (1996). Predatory Tactics: Survival in the Microcosmos. NTSC videocassette, 42 minutes. Cytographics, Ascot Vale, Australia.Ref.: http://www1.umn.edu/news/prod/groups/ur/@pub
/@ur/documents/asset/ur_87651.jpg
Actinophrys
Chlamydomonas
axopodia
flagella
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finis
- Example: 3 um/15 min. = 0.2 um/min = 2 x 10^-7 m/min.one 9+2 axoneme = 233 protofilaments/flagellaone tubulin monomer/4 x 10^-9 m2 x 10^-7 m/min. x one tubulin monomer/4 x 10^-9 m = 50 monomers/min.50 monomers/min. x 233 protofilaments/flagella = 11,650 monomers/flagella min.11,650 monomers/flagella min. x 2 flagella/cell = 23,300 monomers/cell min.
- 2. Assuming de novo tubulin synthesis, calculate the number of amino acids polymerized per minute to make this tubulin.23,300 monomers/ min. x 450 aa/monomer = 1.05 x 10^7 aa/min.
- 3. Assuming that all mRNA for tubulin synthesis is being made de novo and that each mRNA is translated only once, calculate the number of RNA bases transcribed per minute. What if every mRNA were translated 100 times?1.05 x 10^7 aa/min. x 3 bases/aa = 3.15 x 10^7 bases/min. for single use mRNA3.15 x 10^7 bases/min./100 = 3.15 x 10^5 bases/min. for 100x use mRNA
- 4. Assuming that every mRNA is translated 100 times, and assuming that RNA Polymerase II can polymerize 2,500 bases per minute at maximum, how many tubulin genes are present in the Chlamydomonas reinhardi genome?3.15 x 10^5 bases/min. //2.5 x 10^3 bases/min/RNA Pol II = 126 genes