BIOC1 Final exam - practicals PDF

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Practical A – Molecular BiologyHuman DNA - If the DNA from a single human cell were unravelled it would be approx. 2 meters long - Genes are typically around 104 base pairs long, spread throughout the genome of large regions of non-coding DNA in between. Bacterial DNA - DNA is much less than Eukaryo...

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Practical A – Molecular Biology Human DNA - If the DNA from a single human cell were unravelled it would be approx. 2 meters long - Genes are typically around 104 base pairs long, spread throughout the genome of large regions of non-coding DNA in between. Bacterial DNA - DNA is much less than Eukaryotic, also difference in arrangement of DNA in these cells, a single molecule of genomic DNA and plasmids – circular molecules in the cytoplasm, often carrying antibiotic resistance genes. Circular DNA found within bacteria is supercoiled – when a double stranded DNA helix (called a replaced molecules) is twisted around itself. This is more compact than a relaxed open circle, which affects its migration during electrophoresis.

Both supercoiled and relaxed can be cleaved by enzymes to give the same linear molecule. Restriction endonucleases - Made naturally by bacteria as a defence against viral infection; cutting up viral DNA to ‘restrict’ their growth. And bacterium protects its own DNA by methylating the target sequences. - Don’t act randomly, recognise specific short sequences in the DNA and cut at those points, examples include:

All restriction sites are palindromic, the frequency with which the sites occur is related directly to the number of bases recognised. n 1 - Probability for n particular bases occurring in a sequence is ( ) 4 - Short restriction sites are more likely to occur randomly in DNA. So short recognition sequences will cleave any random piece of DNA more often resulting in smaller fragments of DNA, lengths identifies by gel electroph.

Gel electrophoresis; setting up an Agarose gel separation and observing fragments on a transilluminator - Phosphates on DNA have –ve charge at pH above 1, so they will migrate towards the anode. - Agarose is a polysaccharide, DNA molecules move through the buffer-filled pores in the gel, probably move as ‘rods’ - The larger the fragment, the more drag it will experience - Migration is inversely proportional to molecular size. Smaller fragments move more quickly Ethidium bromide is used to visualise DNA, it intercalates between bases, takes on a pink-orange fluorescence when illuminated with U.V. light at 300nm Restriction mapping Of a DNA fragment shows the portions of the different target sites for various restriction enzymes. Many restriction sites – problem, computer analysis is needed. 9kb circular plasmid, digested w enzyme P = linear 9kb, - only 1 restriction site Analyse plasmid for locations of restriction sites for 2 more enzymes – double digest; - P + Q => two fragments of 2 and 7kb, so Q cut once, 2kb away from the P site - P+R => 2 fragments, 3 and 6kb so there’s only 1 restriction site for R - Info on how restriction sites are positions in respect to P site, 2 possible theoretical plasmids;

Procedure 1. Mix enzymes needed, design with 1ul added, and 1ul of restriction buffer. 2. Mix contents by vortexing 3. Spin in a microfuge – bottom sediment 4. Incubate for 30 mins at 37oC 5. Arrange buffer plants, well comb; Pour 50ml of molten Agarose into apparatus, leave to set then remove well comb 6. Add TBE – contains ethidium bromide 7. Add 3ul of loading due to the tubes 8. Load all digests and DNA markers 9. Connect to a power supply at 70V for approx.. 40mins 10. Disconnect and photography gel Analysis - Measure the distance in mm that each fragment in the marker has migrated from the well. - Plot aa standard curve of: x = log10 number of base pairs y= distance migrated - Draw a line of best fit, linear although larger deviations with larger fragments, lot of scatter here - Use this as a calibration to calculate the length of base pairs in the digest - To get actual number of base pairs for 10^logscale number Tips: - Measure from the bottom of the band – furthest it’s gone - Always cross check the bp number to be sure - All about comparing and picturing in mind. From photo; - M – marker lane - P = non-recombinant vector plasmid - 1-x = potential recombinant plasmids Errors between the two methods; due to estimations and ruler error Large error in measuring smaller fragments. Distances measured by larger fragments show a larger gradient vol of solute ×100 col of solution Glycerol is present in the loading buffer to make the sample denser than the running buffer so it sinks to the bottom of the gel % solution=

Practical B – Transamination and Deamination -

Animals ingest amino acids, intermediates needed for pt synthesis, must be deaminated, so that the carbon skeletons can be used as metabolic fuels 0 glycogen/fats Interconversions of aminoacids and oxo/keto acids, to channel the amino nitrogen into urea for excretion

Transaminases Transaminases have pyridoxal phosphate as a tightly bound cofactor, derived from B6. Keq is close to 1, they move amino groups from an a-amino acid, to a 2-oxoacids, exchanging amino and ox-acid structure - Glutamate amino transferase

Funnelling of amino acids to aspartate (urea cycle) or glutamate (ox. Deamination) Deamination - Glutamate dehydrogenase – oxidoreductase (deamination) oxidative deamination of glutamate -

Glutamate dehydrogenase will use either NAP+ or NADP+ as a coenzyme. The toxic ammonium ion NH4+ is transported as glutamine to the liver where it is converted to carbamoyl phosphate, enters the urea cycle Heart extract; supernatant dialysed exhaustively; many changes of dialysis buffered against ice-cold phosphate buffer TLC; separation due to properties, usually polarity The a.a that travels the shortest distance is the a.a at the N-terminus Identifying products:

The 2,4-dinitrophenyl-hydrazones of the oxoacids are yellow and can be seen directly The amino acids are detected by spraying ninhydrin and heating, purple colour Procedure: 1. Add extracts, incubate for 30 mins at 37oC 2. Add 1.5ml ethanol to amino acid reagents and 1.5ml of2, 4 dinitrophenylhydrasine to oxo-acids 3. Put oxo in water bath for 5 mins, and then add 0.5 ethyl acetate. 4. Centrifuge all tubes, 5 mins at ¾ speed 5. Spot the samples on TLC plates Solvent systems; - Amino acids= ethanol: aq ammonia conc 70:30 - Oxoacids= n-butanol:ethanol: aq ammonia conc, 70:10:20 Run the plate until solvent front is near the top  Unreacted dinitrophenylhydrazine runs irregularly near the top, so ignore the top spots pyruvate DNP often produces 2 spots close together (isomers)  Ninhydrin creates a purple spot as; Ninhydrin reacts with the amino group of the free amino acid which upon further decarboxylation and removal of an aldehyde produces an adduct which forms a purple complex with another molecule of ninhydrin. tube pyruvate Glutamate Alanin oxogl NAD Transam Dehyd e u +(R) r R S R S 1 5 2 6 3 7 4 8 Transamination reactions; Tube 3; pyruvate and glutamate, produces alanine, so glutamate and alanine spots corresponding to A +2 Tube 7: pyruvate and glutamate, produces oxoglutarate, corresponding to 5 and 9 Deamination reactions Tube 4: Glutamate reacts to form oxoglutaate as NAD+ is provided. Glutamate spot but oxoglutarate spot isn’t visible Tube 8: glutamate reacts to form oxoglutarate as NAD_ provided, small oxoglutarate spot seen and unreacted DNP indicates very little product PURUVATE + GLUTAMATE = ALANINE + OXOGLUTARATE

Dialysis calculations - Removes small molecules by allowing their diff through a semi-permeable membrane whilst retaining the large mol in the dialysis bag. - Protein (a.a.) of 10mM in a dialysis bag od 10mL is placed in a 1L buffer for an hour After 1 hr, conc is evenly distributed throughout the whole system, this is starting vol c× total vol - Done three times, what is the conc of a.a. In the bag at the end 1 dialysis: 10 =99 × 10−6 M ( 10 ×10−3) × 1010 2nd and 3rd – carry on the conc, same fraction st

Practical C – Cell Biology Use an arbitrary (graticule in the eyepiece) scale that must be calibrated for the fixed and stained samples. This is aligned with a second calibration slide, precisely 1mm in length, marked in 100 divisions. - Sale must be calibrated for other objectives, calculating the numeric relationship [p; 1 a.u. = A mm - Make measurements on the arbitrary into meaningful units of measurement Haemocytometer is 1mm2, must identify the part the cell occupies as different areas are scaled differently -

Analysis - Explain why it is what it is, consider how the size might relate to function of the cell being investigated, any relationship? - Is there a difference between plant and animal cells? - How do unicellular sizes differ> why? - What limitations are there to cell size in each case? Must be able to - Calculate accurately the speed of movement and size of selected organisms Setting up a microscope 1. Turn the objective turret to low power (x10) and place the slide on the stage 2. Switch on the light source and bring specimen into focus using the course and fine focus controls 3. Raise the condenser as far as possible, using its focussing controls 4. For unstained living specimens, partially close the sub stage diaphragm to increase contrast 5. Increase magnification by rotating the objective turret to select the appropriate objective, make sure that the lens doesn’t come into contact with the slide. 6. The x100 objective functions only under oil immersion Calibrating the arbitrary scale 1mm of calibration slide has 100x 10um units. Calculate the dimensions for each objective lens by superimposing the two scales *Use the entire length of the shorter scale to complete the calibration in order to reduce experimental error Part 1 Zea mays; transverse section of stem; different diameters due to plant interactions and hormones Skin; cannot see the middle layer due to; connective tissue a mesh of elastin and collage, good vascularisation, and fat cells Bacteria; gram positive = violet cocci, gram negative = pink rods; measure the diameter of a coccus by measuring multiple cells and divide by number of cells in the group Cell motility Movement of cells themselves and organelles or cytoplasm, due to motor proteins, that converts the chemical energy in ATP into mechanical work. Motorproteins include:

Myosin generating movement by its interaction with Actin filaments (MICROFILLAMENTS). Kinesin and dynein generate movement along MICROTUBULES Measuring the speed of; (a) Paramecium; a ciliate, MICROTUBULE-based motility (b) Euglena; protist – euk doesn’t fit into any kingdom (c) Amoeba; protozoan, uses amoeboid movement – ACTIN + MYOSIN (d) Homo sapiens; bipedal metazoan, motion in a range of media. Haemocytometer 0.25mm, smaller division squares are 0.2mm, smallest square is 0.05mm

Practical D – Enzyme Kinetics Enzyme assay 1. pH - Find optimum pH by varying pH and measuring product produced 2. Incubation time for reactions; to find effect of [substrate] on reaction rate, to ensure it’s not running out during the incubations, to see how fast the enzyme is capable of going - How long can an incubation run before the decreasing [sub] starts to have a negative effect on the reaction rate - Over which range of time is a constant rate (linear) of product formation - In the presence of an enzyme it’s best to use as short a time as possible, highest reaction rate – as long as it produces a measurable amount of product. Subsequently, use less time than identified in the exp - not too small though as v small product formed, not accurately measured and experimental error, not accurate - Aim is to get a straight line~10 mins 3. Effect of [enzyme] on reaction rate Use default [substrate] and vary the [enzyme]; to find a small vol of enzyme that still results in a measurable amount of product. Rate increases linearly, then later start to curve and plateau, indicates substrate is running out too quickly with higher [enzyme]. Due to enzymes being expensive we want to use the least enzyme ~100uL 4. Effect of [substrate] on reaction rate Keep the [enzyme] constant, to the low value. Graph will plateau, around 150uL subs, as enzyme is saturated. Plateau on subs conc Vs mol of product formed will tend to Vmax but doesn’t reach it. Focus on the [subs] that gives a linear portion on the graph ~75mM - DRAW A LINEWEAVER BURKE PLOT - Difference in [s] affecting reaction rate (v) - Calculate Vmax and Km Add a trendline and straight-line equation helps Plot a scatter chart, not line graph, plot 1/[S] on x and 1/R on y axis Get a trendline, and forcaste backwards to get y and x intercepts Y intercept = 1/Vmax X intercept = -1/Km =-c/m

Equation sheet:

Practical E: Oxidative phosphorylation ATP synthesis - Oxidative phosphorylation – mitochondria phosphorylate ADP to ATP, this is linked/coupled to the flow of electrons from NADH or FADH2 through a series of electron carriers (the electron transport chain) in the inner mitochondrial membrane. - A pair of electrons moves to different pt complexes and H+ ions are moved across the membrane outside, setting up a membrane potential. This doesn’t happen indefinitely, H+ ion gradient reaches a limiting value and no more H+ can move, here electrons cannot flow are no O2 is used - H+ ion s flow back down their conc gradient through ATP synthase, converting ADP to ATP, ATP synthase is a molecular motor driven by the movement of H+ ions back into the matrix. - The electron transport chain has 4 protein complexes (3/4 are H+ ion pumps, complex 2 isn’t) and 2 mobile electron carriers (Coenzyme Q and cytochrome c) - NadH2 is reoxidised by complex 1. FAD-containing succinate is oxidised by Complex 2. o Succinate is the substrate fewer H+ ions are pumped across, so we get less ATP for each pair of electrons from FADH2 than NADH; NADH = 2.5ATP//FADH2 = 1.5ATP NADH pumps 10 H//FADH2 pumps 6H - Coenzyme Q/Ubiquinone carry electrons from complex1/2 to complex3. Cycles between CoQ and CoQH2 - Cytochrome passes electrons from complex3 to complex 4, cycles between Fe2+ and Fe3+, membrane associated protein - Complex 4 moves H+ and passes electrons to O2 Oxygen Electrode - Measure the oxygen concentrations. O2 is reduced at the cathode and the flow of current tells us the oxygen concentration. A constant polarising voltage is applied across the cell, sufficient enough to reduce O2 CLARKE ELECTRODE: saturated KCl, separated from the test solution by an Oxygen-permeable PTFE membrane. The volume of electrolyte under the membrane is v small but the stirrer ensures the O2 concentration at the electrode is the same as that in the bulk sample, accurate measurement

Respiration States: *Assume there is plenty Pi and substrate* State 2: Mt with substrate and no added ADP, in the absence of ADP, little or no oxidation of substrates and little/no consumption of O2. State3: once ADP is added, the initial rate of O2 use is proportional to the amount of ADP added State 4: ADO is exhausted. The rate of Oxygen consumptions drops to zero

Uncouplers – compounds that can dissolve in the mt membrane and carry H+ back across the membrane. They short-circuit ATPsynthase, H+ gradient doesn’t build up and so electrons flow down to reduce O2, but this isn’t coupled with ATP synthesis. Therefore uncouplers inhibit the coupling between the electron transport and phosphorylation reactions, inhibiting ATP synthesis without affecting the respiratory chain and ATP synthesis - Source of error Calculating P/O P/O = ratio of ADP phosphorylated to O2 atoms used *Cannot calculate P/O ratio in the presence of an uncoupler, if there’s no ADP or oxygen used * Low background rate: in reality, mt turn over a little ATP themselves, and can get a little damaged in preparation. The state 2 and 4 are not zero. Measurement needs to compensate, so we only see the O2 used as a result of the ADP we added 1. Calibrate the O2 electrode by noting the distance between 0% (chemically removed O2) and 100% (O2 conc of air saturated experiment medium) 2. Will state 0-100% represents x nmol of O2 atoms, calculated from the volume and O2 concentration

If more O2 needed for the same ADP, the P/O ratio is lower – case for FADH2 From a graph:

1. 2. 3. 4.

Rule straight lines for state 2, state 4 and one horizontal where ADP was added Point x = where the state 3 and state 4 lines intersect Measure vertical distance between x and state 2 line (58mm) Measure the distance between 100 and 0% saturation = 1000nmol of O (88mm) 5. Oxygen used is 58/88 *1000nmol = 659nmol 6. Mol of ADP added/Oxygen used = 1000nmol/659nmol = 1.52

*If talking about O2, must divide moles of O by two

Experimental conditions 1. Different substrate; malate (NAD+) and succinate (FAD) 2. Vary he amount of ADP 3. Add dinitrophenol, with and without ADP Once all the ADP has been converted to ATP, ATP synthase usually stops working, so the etc is forced to be stopped too. DNP allows the protons

another route into the matrix, allows the etc to pump H+ out as fast as DNP allows them back in, which is much faster. Therefore the function of ATP synthase no longer has any effect on the function of the etc, and the etc will keep on going until the O2 is nearly depleted. 4. Add compounds that will inhibit the overall process with and without DNP; amaytal, rotenone, antimuycinA, oligomycin, potassium cyanide Results: Amytal//Rotenone O2 consumption doesn’t occur with malate, but does with succinate. The only difference between the two substrates is that succinate is oxidised on complex 2, while NADH from malate oxidation is oxidised on Complex 1. Amatyl prevents the etc from using the NADH from malate, so it must inhibit the complex 1 Antimycin A//Potassium Cyanide Inhibits the electron transport chain for both substrates, so must inhibit after complex1/2. This is inhibited even in the presence of DNP, so it can’t be anything to do with ATP synthase; it is inhibiting Complex 3 or 4 Oligomycin State 2 or state3 respiration is always observed, indicating that all components of the etc are at least working, DNP allows O2 at a high rate, so what is holding the etc back is the re-entry of H+. Therefore oligomycin must be inhibiting ATP synthase...