Lab 2
Flagella Regeneration
in Chlamydomonas reinhardtii

or how simple microscopy can be used to calculate rates of assembly of macromolecules into cellular structures

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

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

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?

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

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