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Biology Final

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Prepared 3 beakers with contents listed below. ( a. Beaker 1: glucose only b. Beaker 2: Starch only c. Beaker 3: Starch + amylase). Poured contents of each beaker into its respective fermentation tube, ensuring the tail portion of the tube was filled with liquid. Placed tubes in an incubator at 37 degrees, measuring distance between tip of tube tail to fluid level at 20, 40, and 60 minute intervals. Calculated gas volume using this distance along with radius of tube tail. Why is alcohol fermentation needed in yeast to carry out glycolysis In an environment absent of oxygen, yeast undergoes alcohol fermentation in order to continue ATP production but also to recycle NAD+ that is needed for glycolysis.

Without this recycling, glycolysis and therefore cellular respiration cannot proceed and organism will die.

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o How efficient is fermentation- 2% (2 ATP) efficient compared to aerobic respiration, which is 39% (36 ATP) efficient at capturing the energy released in the form of ATP o What amylase does to starch- Breaks down starch to individual glucose units o What gas is produced- CO2 The experiment 1.

How do we take advantage of the gas produced to measure fermentation rates Measure the distance between the top of the tube tail to the fluid level, use this distance to calculate the volume of CO2 gas produced at each timed interval (also using radius of tube tail). 2. What effect is temperature expected to have on this system Enzyme (amylase) works best at its optimal temperature thereby reducing the activation energy, allowing the reaction to go faster. 3.

What is the point of performing treatment 2 (starch) versus treatment 3 (starch + amylase) -To determine whether fermentation will happen while sugar is in its complex, polymer form or only once broken down to its monomer units (glucose). Beaker 3 introduces the enzyme that breaks starch down to its monomer units. 4. What are the hypotheses for the 3 treatments and why? Beaker 1 – CO2 will be produced the fastest. Sugar is already at the monomer unit Beaker 2 – CO2 will not be produced or will be the slowest rate of all.

Starch needs to be broken down to glucose in order for fermentation to occur. There is no enzyme in this beaker to break down the starch. Beaker 3 – CO2 will be produced the second fastest. As amylase is in this beaker, starch can be broken down to its monomer unit but it will not be as fast as Beaker 1 since the latter already contains its monomer units. 5. Given the data, be able to graph the data and discuss whether the data match the experimental predictions (without doing statistical analysis) This is at the discretion of the instructor.

May be a line graph or bar graph, charting volume at each interval, or charting the total volume after 1 hour, or something else B. Photosynthesis experiment o Gross versus net photosynthesis Gross – total products produced from the conversion of light to chemical energy (i. e. total sugar, O2, etc. ) Net – total defined above minus the amount of products cycled back for respiration o Overall gas exchange of a plant

As photosynthesis occurs, leaves take in CO2 from the bicarbonate solution and produce O2 gas, which displaces the infiltration liquid currently filling the intracellular spaces of the leaf disks o Relationship between light levels and gas exchange in plants In this lab the treatment with the higher light intensity produced the higher rate of photosynthesis, faster rates of CO2 was consumption, and O2 production, compared with disks under the room intensity light o The basic process used to measure photosynthesis 1. Why infiltrate the leaf discs

Flood the intracellular spaces with liquid and flush out the gas/air, which will allow the leaves to sink in the bicarbonate solution 2. Why do the leaf discs sink or float: relate to leaf anatomy and physiology They sink because the mesophyll space is filled with infiltration liquid (heavier/higher specific gravity) rather than air (lighter/lower specific gravity), which allows them to sink. As photosynthesis occurs, the O2 produced displaces the liquid from the mesophyll space thereby allowing the leaf disks to float. 3.

How that relates to the rate of photosynthesis The faster the leaf disks float, the faster the rate of photosynthesis. The slower they float, the slower the photosynthesis rate. 4. Where is the source of CO2 for the leaf disks Bicarbonate solution C. T-tests o Be able to use raw data to calculate means and standard deviations and t-values and to interpret the t-values and p-values to determine whether your hypotheses are supported Review pages 17 – 23 in lab manual § _You will be provided with the formulas and the t-table

D. Starch prints o The logic behind the process By manipulating light exposure on leaves (i. e. via black and white negative) and therefore areas where photosynthesis and starch production occur at the highest rate, one can develop a picture image on a leaf from the black and white negative. o How a negative image is turned into a positive image inside the leaf tissue Areas of the leaf where light was exposed were able to go through the process of photosynthesis (and carbon fixation), thereby generating starch.

Areas that did not receive light did not produce starch. When the leaves were boiled, bleached then stained in iodine to identify areas of starch production, areas where starch was produced were the darker areas and areas where little or no starch was produced were the lighter areas, in fact developing the positive image. o Relationship between light levels and rate of photosynthesis / starch production Again the more intense the light (without damaging the leaves of course), the greater the rate of photosynthesis, the more starch produced. What the experiment tells us about transport of starch Parts of the leaf exposed to the light source were able to photosynthesize, and therefore able to produce starch.

D. Satellite DNA o What is satellite DNA- Highly repetitive DNA not known to contain any genes o What is a bp (base pair) The complementary purine and pyrimidine base pairs found on double-stranded nucleic acid chains. Example, adenine and tyrosine are a base pair o Coding versus non-coding regions Coding regions code for proteins.

Non-coding regions do not o What are restriction enzymes Enzymes that cut nucleic acids at specific sequences into smaller nucleotide units 1. Significance of digestion time The more time a restriction enzyme has to digest the DNA, the more restriction fragments of same length will be produced with the targeted repeating sequences. o What we need to do (in general terms) to get the nuclear DNA Review 72-75 of lab book o What is gel electrophoresis and how it works 1. The role of the electrical current

Helps the DNA fragments move through the gel. As DNA is negatively charged, it will move from negative electrode to positive electrode. Repelling the negative charge end helps it move down the gel to the positively charged end where it’s more attracted 2. The role of the gel material Gel acts as a sort of filter, helping to separate the DNA strands of specific length 3. The role of the dye Helps make DNA samples in the wells visible as the samples move through the gel 4. The role of the stain

Helps the bands appear as more prominent lines in the gel o Why do we sometimes see a smear of DNA as opposed to discreet bands on a gel May be too much DNA loaded in the gel well or there might have been protein and RNA contaminating the DNA mixture or cuts were more random such that you have a jumbling of different length fragments o How to find the size of the DNA fragments showing up on a gel- Compare the bands of sample DNA to the bands of DNA standard where lengths are known

Cite this Biology Final

Biology Final. (2016, Oct 02). Retrieved from https://graduateway.com/biology-final/

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