Lab 5 Alcohol Fermentation & Aerobic Respiration PDF
Document Details
Uploaded by Deleted User
Tags
Summary
This document details lab objectives related to cellular respiration and fermentation. It explains the basic differences between the two processes, the rate of gas produced during fermentation, and similarities between carbohydrates as fermentation substrates. A detailed understanding of the processes of photosynthesis and respiration is also a core part of the document.
Full Transcript
Lab 5 Alcohol Fermentation & Aerobic Respiration Understand the basic differences between the processes of cellular respiration & fermentation, including their overall rates he main difference between cellular respiration and fermentation is that cellular respiration T requires oxygen to...
Lab 5 Alcohol Fermentation & Aerobic Respiration Understand the basic differences between the processes of cellular respiration & fermentation, including their overall rates he main difference between cellular respiration and fermentation is that cellular respiration T requires oxygen to proceed. C6 H1 2O6 + 6 O2─ ─────> 6 CO2 + 6 H2O + ENERGY (36 or38 ATP) + Heat Fermentation: 6 H1 2O6 ──────> 2 C2H5 OH (ethanol) + 2 CO2 + ENERGY(2 ATP) + Heat (occurs for e.g. C in yeast cells 6 H1 2O6 ──────> 2 C3H6 O3(lactic acid)+ ENERGY (2ATP) + Heat (occurs for e.g. in human C muscle cells.) ermentation is less efficient than cellular respiration. It therefore consumes more glucose to F sustain the same energy usage (higher rate). Given the Rate of gas produced during fermentation and the concentration of a yeast solution be able to calculate the specific rate (gas per minute/gram yeast) If you have the concentration and the volume of yeast solution used, you can find the mass of yeast used. Divide the rate of gas produced (Kpa/sec) by the weight of yeast and multiply by 60s to get the rate of gas produced in Kpa/minute/gram yeast. Explain similarities or differences between common carbohydrates as fermentation substrates in yeasts Glucose:can be metabolized by yeasts immediately,used as our baseline (percentage of glucose fermentation :100%) Sucrose:the yeast used in this experiment can metabolizeglucose and fructose immediately. It has the enzyme sucrase that breaks down sucrose into glucose and fructose. The rate is expected to be very close to the one for glucose. Maltose:The yeast can ferment maltose but with a delay due to induction. The rate directly after maltose is added will be very slow. However, after 15 min, which is enough time for the synthesis of transport proteins and enzymes, the rate will be much higher. Lactose:disaccharide (glucose and galactose) yeastused in the experiment does not produce the enzymes to metabolize lactose. The rate will be slow or equal to zero. Water:serves as the control: technically the rateshould be equal to zero.(no carbohydrate) Explain the delay in activity with maltose The few molecules of maltose that enter the cell do so by non-specific transport mechanisms. However, once they do enter the cell, their presence stimulates the synthesis – through transcription & translation – of the enzymes maltose permease & maltase. This then allows the more rapid entry of maltose into the cell. The lag period for the appearance of these enzymes is yeast strain-dependent. In the end, maltose is hydrolyzed into two glucose molecules, which are then fermented into ethanol. Discuss the relevance of the R.Q. ratio (respiratory quotient) in terms of types of substrates being used (carbohydrates or fats) in cellular respiration and be able to calculate the respiratory quotient of a given substrate The R.Q. (respiratory quotient) of a tissue is defined as the amount of CO2 production during respiration divided by the amount of O2 uptake. It can be used to determine the contribution of different tissues in energy production (can find out which one is being metabolized). Since carbohydrates are always a CH2O ratio, they usually have a RQ of 1. Fats and proteins usually have a lower R.Q. They produce less CO2 for the same amount of O2 used. Describe the set-up for the measurement of the rate of aerobic respiration in pea seedlings (why was the soda lime necessary?) To measure the rate of aerobic respiration, we measure the oxygen uptake using a pressure sensor. Since, the volume of O2 taken up should be the same as the volume of CO2 released, an inch of soda lime over the pea seedling is used to absorb the CO2 gas produced during respiration. The tube is wrapped with aluminum foil. The aluminum foil is used to prevent chlorophyll from carrying out photosynthesis, which could interfere with measurements of respiration. Lab 6 Structure and Function of Leaves Know the general structure of plant cells and estimate their size. Know the structures of different types of leaves Cuticle: waxy layer covering the upper and lower surface of the leaf, prevents water evaporation/loss, CO2 can’t pass through it. Epidermis: upper and lower layer that acts as a barrier. Vascular bundle: the xylem (upper portion, thick-walled cells, stained red, conducts water and minerals in only the upward direction) and the phloem (lower portion, small, thin-walled cells, stained green, conducts food/organic molecules in both directions). These are continuous with the same tissues in the stem & the root. Mesophyll: Primary photosynthetic tissue of the plant. Many chloroplasts are located within individual mesophyll cells. The palisade mesophyll: cylindrical cells nearest to the upper epidermal layer The spongy mesophyll: containing irregularly shaped cells beneath the palisade layer, & having many intercellular spaces. Stoma: made up of 2 guards cells and the pore (opening), allow gas to enter/leave the cell (co2 enters o2 leaves). However, a lot of water is lost (osmosis goes from high concentration inside the cell to low concentration of water vapor outside) and replaced by upward diffusion of water from lower parts of the plant. This whole process is referred to as transpiration. Understand how these structures reflect the adaptation of plants to different environments with respect to: water conservation: The intracellular osmolarity controls the water content of the guard cell and therefore its shape. When guard cell osmolarity becomes greater than the extracellular fluid, water enters the cell via osmosis causing turgidity and stomatal opening. When intracellular osmolarity becomes lesser than the extracellular fluid and if the water lost in transpiration is greater than that replenished by roots (as may occur in dry conditions) water leaves the cell causing flaccidity and stomatal closure. The stomata are also regulated by exposure to light in the photosynthetic spectrum, low CO2 concentrations, and high humidity. gas exchanges: The stomata (singular stoma) are the systems made up of pores and guard cells through which gas exchange (CO2 in, O2 out) occurs and are usually located on the lower surface of the leaf. Gas exchange in woody stems is accomplished partly through exchanges with vascular tissues, and partly by means of lenticels, loose groups of cells in the otherwise impervious epidermis of corky layers of a woody stem. light absorption be able to identify a picture of a slide of a mesophyte, hydrophyte or xerophyte leaf and describe specific adaptations of these leaf types Xerophyte: adapted to dry habitats: prominent cuticle, multi-layered epidermis, stomata located on the lower surface in specialized pits lined with epidermal hairs. Hydrophyte: adapted for life on water. the stomata in the upper epidermis only, the very thin cuticle, the presence of chloroplasts in the epidermis, & the very large air chambers in the spongy mesophyll. Mesophyte: flat leaves with a moderate cuticle and evenly distributed stomata. The leaf will typically have a normal size and shape. Understand the difference between absorption and reflection of light, and that the light that is absorbed “powers” photosynthesis; Observe, graph and understand the absorption spectra of different plant pigments; Pigments, or chromophores, are substances which absorb visible light of certain wavelengths (colors) while reflecting or transmitting light of other wavelengths. The wavelength that is not absorbed is the one we see. If we observe the absorption spectra of these different plant pigments, Chlorophyll A and the accessory pigments carotenoids ( xanthophylls 1-2(yellow) and carotenes(orange)) and chlorophyll B, we can see that there’s no absorption at the wavelength corresponding to the color green. In plants, the energy absorbed contained in photons (light energy) by the green chlorophylls & yellow/orange carotenoids can be "trapped" & used to synthesize carbohydrates & other food molecules. In addition, carotenoids photoprotect: they absorb and then dissipate excessive light energy that would damage the leaf and other photosynthetic parts of the plant. In doing so, they act as antioxidants. Understand how the difference in the solubility of the different plant pigments in a particular solvent (mobile phase), observed using chromatography is linked to the molecular structure of these pigments Using chromatography, as the eluent (mobile phase) moves through the cellulose in paper by capillary action, the different pigments are separated based on their solubilities in a particular solvent (who is less polar) and their attraction to a particular solid material (usually polar). The most nonpolar molecules in the mixture will spend the most time in the mobile phase and travel the furthest over time (faster rate), while the most polar molecules will spend the most time in the stationary phase and travel the least over time slower rate). Chlorophylls A-B: polar ring structure and nonpolar tail: CHO in B and CH3 in A. Both won’t travel very far but B will be the slowest due since it’s more polar than A. Xanthophylls: long hydrocarbon chain but with an OH group attached Carotene: long hydrocarbon chain. will travel the furthest. Lab 7 Restriction Enzyme Digest and Electrophoretic Analysis of the Recombinant DNA Plasmid pGLO Describe restriction enzymes, specifically endonucleases, and describe their protective roles in the organisms that produce them. Restriction enzymes are cellular enzymes in bacteria that cut up DNA that is identified as foreign by the bacteria. These cells use restriction enzymes to defend themselves from bacteriophages, a type of virus that infect bacteria by injecting nucleic acids (DNA or RNA) into the bacterial cell. Restriction endonucleases, a type of nucleases that function in breaking the phosphodiester bonds that link adjacent nucleotide monomers in DNA, fragment viral DNA at internal positions as soon as it enters the bacterial cell. Explain why restriction endonucleases do not cleave a bacterial cell's own DNA as well as that of the viruses that are infecting it. Methylation. Bacteria modify their own DNA, using other enzymes known as methylases to add methyl (CH3) groups to some of the nucleotides in the bacterial DNA.Restriction endonucleases cannot bind to Dna sequences that have been methylated. Describe how restriction enzymes can be used in recombinant DNA technology; Restriction endonucleases first recognized in bacteria are now essential tools in biotechnologies such as the production of recombinant DNA, cloning and the mapping (analysis of DNA sequences) of genes and entire genomes in both prokaryotes and eukaryotes. Since the DNA code is the same for all organisms, and many restriction endonucleases cleave their palindromic recognition sequences asymmetrically, leaving a single-stranded overhang at each side of the cut, called a sticky end, we can rejoin two different strands cut by the same restriction enzyme using DNA ligase. Use a restriction digest map of a plasmid to calculate the size of the fragments generated; Understand how DNA sequences are separated using agarose gel electrophoresis; In such a technique, the samples containing the pGLO plasmid, the restriction enzyme(s) and the dye as well as the DNA ladder are placed in each well next to the anode. The electrophoresis is run at 80-120V for around 45 min. Size is the only important factor (not charge bcs all the DNA fragments are negatively charged). Small molecules placed in the gel move easily through the small pores in the gel lattice, while large molecules have more difficulty. Therefore, the smaller fragments are gonna be closer to the positive electrode. Identify fragments by size in agarose gel electrophoresis. Lab 8 Bacterial Transformation Expression of GFP in Bacteria Describe transgenesis and briefly comment on its use in biological research, biotechnology, agriculture, and medicine using examples. It’s the process of introducing a gene or genetic material from one organism into the genome of another organism, often of a different species, in order to change the organism’s trait. In agriculture, genes coding for traits such as frost, pest, or spoilage resistance can be inserted into plants. In medicine, gene therapy treats diseases caused by defective genes by inserting healthy copies of the defective gene in a sick person’s cells. Although bacteria reproduce asexually, these organisms often transfer DNA from one cell to another within a population (and sometimes between species) by a process called horizontal gene transfer (HGT). Describe the procedure used in a CaCl2 bacterial transformation and explain the function of each step. Main goal: insert the plasmid DNa into the e.coli cell and provide an environment for the cells to produce their newly acquired genes. Step 1: Make the cells artificially competent usingCaCl2, which neutralizes the repulsive negative charges of the phosphate backbone of the DNA and the phospholipids of the cell membrane to allow the DNA to enter the cells during the heat shock procedure. Step 2:Use heat shock to increase the bacterial uptakeof foreign plasmid DNA by increasing the permeability of the cell membrane to DNA. The rapid temperature change (ice, hot water, ice) has to be timed very precisely because we don’t want to kill the cells either. Step 3: Provide them with nutrients (LB broth) and a short incubation period to begin expressing their newly acquired genes. It allows the transformed cells to recover from the harsh heat shock but also to make the ampicillin resistance protein beta-lactamase. Describe the general features of a plasmid vector and the specific elements of pGLO and Explain what an operon is and how the pGLO plasmid was engineered Operons are groups of related genes that are often clustered together and transcribed into RNA from one single promoter. Promoter is a nucleotide sequence in DNA that is the binding site of RNA polymerase, positioning the RNA polymerase to begin transcription at the appropriate position. It is usually found upstream of the coding region of a gene. Recombinant plasmids, such as pGLO, were made using restriction enzymes as tools to piece together the elements of the system. It has a gene for antibiotic resistance, a gene for green fluorescent proteins and one for gene regulation (the arabinose operon). The arabinose operon (araB, araA and araD + controlled by a promoter) is controlling the gene for GFP, which can be switched on by adding the sugar arabinose. When arabinose is present in the environment, bacteria take it up. Once inside, the arabinose interacts directly with arabinose operon and the interaction causes the transcription of the three digestive enzyme genes, which break down arabinose who eventually runs out. In the experiment, the genes that code for the 3 genes that break down arabinose have been replaced by a single gene that codes for GFP. What advantage would there be for an organism to be able to turn on or off particular genes? Inducible genes prevent wasteful overproduction of unneeded proteins and allow adaptation to different conditions. Good examples of highly regulated genes are the enzymes which break down carbohydrate food sources. If the sugar arabinose is present in the growth medium it is beneficial for bacteria to produce the enzymes necessary to catabolize the sugar source. Conversely, if arabinose is not present in the nutrient media, it would be very energetically wasteful to produce the enzymes to break down arabinose. List the control treatments used and explain the purpose of each of these trials. The -Pglo in the LB plate gives the following information about the experiment: -This control was to confirm that the heat shock process did not affect the growth of the bacteria, -This control was to confirm that the CaCl2 transformation solution did not affect the growth of the bacteria -This control was to confirm that the transfer of the bacteria with the sterile loop did not affect the growth of the bacteria, -This control was to confirm that incubation with LB broth did not affect the growth of the bacteria, -This control was to confirm that incubation at 37°C for 20 hours did not affect the growth of the bacteria Analyze the results (actual or hypothetical) of a pGLO transformation experiment and state your conclusions. +PGLO: B/ampicillin/arabinose: expected to see only the transformed bacteria due to the presence of L ampicillin and due to the presence of arabinose, we should be able to see green fluorescent protein under UV light. B/ampicillin: expected to see only the transformed bacteria due to the presence of ampicillin. L No arabinose so no GFP. -PGLO: B/ampicillin: No bacteria should be found due to the presence of ampicillin when no bacteria L should have the gene for antibiotic resistance. LB: bacteria should grow as normal. CONTROL Notes taken during class: w ater control used in fermentation: no other source of CO2 lactose –> yeast lacks enzymes as carbohydrate not in natural environment sucrose is fructose and glucose (fructose in glycolysis as FG6), will digest at the same rate as glucose even though it’s a disaccharide because enzymes are catalyzers peas experiment consumption of oxygen measured, if no aluminium then oxygen p roduced by photosynthesis messes up pressure sensors (won’t be accurate) absorb co2 by soda lime to see pressure change (if no soda lime, CO2 and O2 ratio is 1:1 so no data can be concluded) understand leaves adaptations (based on their environments) -> mesophytes, hydrophytes, xerophytes absorption vs action spectrum (action spectrum includes wavelengths of all pigments, absorption focuses on one pigment’s wavelengths separation of pigments based on how well it’s dissolved (nonpolar solvent), Rf is a measure of how well it’s dissolved (based on how far it gets from baseline compared to the solvent alone) draw cut plasmids and where they would appear in gel (with ladder) endonucleases are derived by bacteria (E coli, etc.) -> protection against viral infections bacterial methylation is why outsider DNA is attacked by restriction endonucleases relaxed plasmid slightly slower than cut linear (imagine folded elastic) coiled plasmid faster than cut linear (imagine tiny scrunched up elastic) 2 cuts of same size –> one line in gel, won’t be thicker (smaller fragments go further and get closer to cathode) DNA fragments are negative 3 components of pGLO: arabinose operon, GFP and ampicillin resistance Control used to see if plasmid is truly expressed and there’s nothing else affecting bacterial growth -pGLO/LB (Lawn, if colonies then something else present) and -pGLO/LB/amp (nothing as OG E coli is not amp resistant) Why not all transformed in +pGLO? Gene just didn’t take in all of them, could be due to particular bacteria heat shock resistance Same number of colonies on both +pGLO plates as they’re derived from one transformed bacteria on each (mother) Why can’t +pGLO/amp glow? No arabinose present -> arabinose activates promoter (transcription factor for GFP) Operon = cluster of genes under the same promoter Inducible gene advantage = doesn’t waste energy on endergonic protein production, adaptation in different environments Changed BAD genes that originally transcribe of enzymes to breakdown arabinose to transcribe GFP instead -> why bacteria glow
