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AS and A Level: Exchange, Transport & Reproduction
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Five tips on presenting and analysing data from core practicals
- 1 Raw data should be presented in a table with clear headings. All table column and row headings should contain units and readings should all be recorded to the same decimal place.
- 2 Raw data should be processed to provide descriptive statistics such as the mean and standard deviation.
- 3 Present data using the relevant graph type. Ensure that you add error bars showing either standard deviation or standard error.
- 4 When describing trends and patterns, manipulate data to calculate the size of key changes. For example, absorbance increases by 0.09 absorbance units between 10◦C and 40 ◦C. Preferably express the change as a percentage increase or decrease. Do not simply quote points, eg at 10 ◦C absorbance was 0.01 and at 40 ◦C it had gone up to 0.1 absorbance units.
- 5 Discuss each phase of the graph. For example if there is a slow increase, followed by a rapid increase, and then the graph levels off and shows a decrease, discuss these four key phases. Do not give detailed descriptions of each small fluctuation. The trends and patterns are the important things.
Meiosis and Mitosis facts
- 1 Meiosis generates gametes/sex cells, whereas mitosis is for growth/repair and generates daughter cells identical to the parent cell.
- 2 During Meiosis chromosome number is halved producing haploid gametes with a single copy of each chromosome. During Mitosis chromosome number is maintained producing diploid daughter cells with maternal and paternal copies of each chromosome, i.e. homologous chromosome pairs.
- 3 Before mitosis and meiosis all chromosomes are copied as part of interphase. At the end of interphase there are two identical copies of every maternal chromosome and every paternal chromosome, so chromosome number has doubled (i.e. at the end of interphase a human cell contains 46 x 2 = 92 chromosomes). The identical copies of chromosomes are referred to as sister chromatids and they are joined by a centromere.
- 4 In meiosis, genetic variation is generated by crossing over during prophase 1, and independent chromosome assortment at metaphase 1 and metaphase 2. During crossing over maternal and paternal chromosomes cross each other, and break at points known as chiasma. Maternal and paternal alleles below the chiasma change places so that the paternal chromosome contains maternal alleles and vice-versa. During metaphase 1 and 2, maternal and paternal chromosomes align randomly on one side of the equator. As the maternal and paternal chromosomes can align on either side, different potential chromosome combinations can occur.
The events in metaphase, anaphase and telophase are identical in both mitosis and meiosis 1 and 2. In mitosis a single division occurs, whereas in meiosis cells undergo 2 meiotic divisions.
a) During meiosis 1, maternal and paternal sister chromatids are separated so that 1 cell contains both maternal sister chromatids of a pair and the other contains both paternal sister chromatids.
b) During the second meiotic division, sister chromatids are separated. The gametes that result contain only 1 chromosome from each pair, i.e. they are haploid.
How to evaluate experimental methods
- 1 When evaluating the reliability of experimental methods, always consider whether all variables other than the independent variable have been adequately controlled. If a variable cannot be controlled has it been monitored to establish any effect it might have?
- 2 All experiments must be repeated to establish reliability. Has the experiment been repeated at least three times? Preferably you should repeat it more than 3 times.
- 3 What does the standard deviation suggest about the spread of the data? If the mean is 5, but the standard deviation is 3, readings vary from the mean considerably. This suggests that the mean does not represent the actual readings.
- 4 How precise are the measurements? If a balance used to measure change in mass only measures to 0.1 g then the reading could be 0.12 g, 0.15g, or 0.18g etc.
- 5 How accurate are the readings? If equipment is re-used for different repeats for example, cross contamination could affect the accuracy of subsequent readings. If tubes are shaken different amounts, different volumes of gas could be released. Consider all potential sources of error and discuss how the procedure could be improved to reduce these sources of inaccuracy.
- Marked by Teachers essays 46
- Peer Reviewed essays 14
At this point the cell wall prevents the cell from bursting and is said to be fully turgid. Turgidity is very important to plants because this is what provides support to make the green parts of plants stand firm. When plant cells are placed in highly concentrated solute solutions they lose water through osmosis. Water is drawn from the cell and the vacuole contracts and shrinks drawing the cell membrane away from the cell wall. The effects of the water loss is called plasmolysis and cells in this condition are said to be plasmolysed or flaccid. When plant cells are placed in a solution which has exactly the same osmotic strength as the cells they are in a state between turgidity and flaccidity and as there is no gradient for the molecules to travel down, equilibrium is set.
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They have small eyes and the mouth is located below the nose. The rats have a touch sensor system that is known as the whiskers on its nasal region called the vibrissa. As for the neck and body, a rat has a short neck with its body consisting of a thorax on its anterior and an abdomen/belly on its posterior. The pectoral region refers to the area where the front legs attach. The ventral surface of the pectoral region consists of the rat's nipples or mammary gland openings. Its tail is long with its anus located on the ventral region of the base of the tail.
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The phospholipids consist of hydrophilic heads which point towards the outside environment and the cytoplasm. The hydrophobic tails repel the water and so point in (see fig 3). Two tails that are highly hydrophobic hydrocarbon chains. Hydrophilic heads (negatively charged) (Fig 3 ref 4) My experiment is based on the fact that proteins/ phospholipids denature when heated. When the membrane proteins are damaged the dye in beetroot leaks out of the cell. This is based on the theory that the more the proteins gain thermal energy the more thermal energy is being converted to kinetic energy.
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Once the isotonic solution is found for each vegetable, its water potential can be found by using a reference graph, which shows the water potential for different concentrations of sucrose. The two vegetables I will use will be a potato and a swede. Swedes are considered a sweet root vegetable and so its sucrose concentration is probably quite different from that of a potato because the sucrose gives the vegetable its sweet taste. Method To carry out this investigation I will need the following apparatus: * * Two boiling tube racks * Ten boiling tubes * 200ml of 1mol sucrose
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For this investigation I will be looking at the affects of osmosis on plant tissue. The cytoplasm of a plant cell and the cell SAP in its vacuole contain salt, sugars and proteins, which effectively reduce the concentration of the free water molecules inside the cell giving a low water potential. The cell wall is freely permeable to water and dissolved substances but it is the cell membrane will acts as the partially permeable membrane. For this experiment the easiest plant cells to use would be that of a potato (chips) and the solution I will use is Sucrose solution.
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Therefore, this means the molecules must be lipid soluble and non-polar, and the beetroot molecules are not. However, facilitated diffusion is the movement of molecules through a membrane by protein channels, and since transport proteins have hydrophilic interior, hydrophilic molecules are able to pass. Facilitated diffusion is the movement of molecules or ions from a region of high concentration to a region low concentration. This results in the molecules travelling down the concentration gradient until equilibrium is reached; where the molecules or ions are evenly distributed. Diffusion is a passive process and it does not require metabolic energy. Diffusion can occur from one part of a fluid to another or across a membrane separating two fluid regions.
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Investigating how changing the concentration of sucrose, affects the time it takes for a reaction to take place with Sucrase3 star(s)
This is shown through the diagram below. As can be seen, there is a point at which the rate of reaction becomes constant, and no longer increases. This is because of the way in which the substrate interacts with the enzyme to form the end products. The diagram above shows that the substrates interact with the enzymes at a specific location. This location is known as the active site. Each enzyme molecule has one active site, and this is where the substrate will attach itself, and the reaction will take place.
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Role of blood in the body and experiment. The aim of this experiment is to either prove or disprove the hypothesis that altitude training increases the bloods efficiency of carrying oxygen around the body.
they carry deoxygenated blood so therefore their walls are thin and made mostly of collagen. they have developed a large lumen so that resistance to flow is reduced. the exception is the pulmonary vein which carries oxygenated blood from the lungs back to the heart where it can be pumped around the body. What is the function and structure of blood? The function of blood is to carry substances around the body (see table below). Blood is the substance that is being investigated, it consists of: * Plasma - this makes up 55% of the blood and is the liquid part of the blood.
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Biological membranes are bilipid layers. Phospholipids create a spherical three dimensional lipid bilayer shell around the cell. Each of the phospholipids molecules has two parts: ? The circle or head is the negatively charged phosphate group which is has a strong polarity because of sharing of electrons within this part of molecule is not quite even. One end becomes slightly positive and other end is negative. So this makes the phosphate head attract other polar molecule- the water. It is hydrophilic.
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The results will show how well the enzyme functions in the varying levels of pH. Purpose The purpose of this lab is to investigate the effect pH has on the efficiency of an enzyme, and to find the optimal pH range of a specific enzyme (meat tenderizer). Hypothesis The percent absorption of the light will gradually go up as the pH rises from very acidic, the enzyme function will peak at its optimal pH, and then begin to decline as the pH becomes higher. It is hypothesized that the optimal pH will be a neutral one, at 7.
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- 1 cm in length Variables that needs to be controlled (independent variable) The variables that needs top be controlled under experimental procedure are as follows: > The amount of distilled water - 5cm3 > The temperature of the water ---- 00C, 100C, 200C, 300C, 400C, 500C,600C, 700C Dependant variable: 1. Amount of die solution. 2. Surface area of the beetroot. Statement of fairness: We are going to investigate the effect temperature has on the permeability of the membrane and there by transporting substances across the membrane. And also we will look the method of transport of substances around the body.
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is converted into alcohol. The actual process occurs when sugars are converted into alcohol using yeast during anaerobic respiration, meaning it does not require oxygen. Firstly, the yeast metabolises the sugars from the original extracted grains. The sugar is then the substrate that breaks down the wort into ethanol, producing carbon dioxide. The word equation for this reaction is: Sugar (glucose or fructose) -> alcohol (ethanol) + carbon dioxide (+ energy) Ref 8 Yeast is a microorganism and is responsible for the different types of beer like ale or lager.
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It would seem that women's presence in sporting environment is modest. This clear distinction between males and females in sport is attributed to the historical notion of men and women within society. Sport is well described as a physical competitive activity that is designed to reveal the physical excellence of men - entertain upper class. It is widely believed that women are biologically and genetically manipulated to care and not to compete. Thus women are socially expected to comply to certain commitments e.g. take care of kids/ house... I think these defaults expectations do influence the lack of women participation in sport.
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As substances dissolve in the water, the water potential drops and becomes negative. Water moves by osmosis to the region with the lowest water potential. When a cell is placed in a hypotonic solution, one that is less concentrated (higher water potential) the solution enters the cell therefore decreasing the water potential. However if a cell is placed in a hypertonic solution, one that is more concentrated (lower water potential) the water leaves the cell. If a cell is placed in a isotonic solution, solutions on either side of membrane have equal concentration, there is not water movement.
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Pilot hypothesis: I hypothesise that the cress will grow to a greater maximum height under blue and red light than they will under green light. I also hypothesise that the greatest heights (on average) will occur under blue and/or red light not green light. Results: Plant Number: Blue Film: (height in mm) Green Film: (height in mm) Red Film: (height in mm) 1 23 3 (no leaves) 15 2 20 0 12 3 20 0 17 4 18 0 16 5 25 0 14 6 27 0 16 7 24 0 5 8 29 0 6 9 26 0 0
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There are different types of simple epithelia, they are: 1) Simple Squamous epithelium - these cells are very flat with each nucleus forming a lump in the centre. The word squamous means scaly to signify the flatness. The cells fit closely together rather like crazy paving. Clearly such delicate thin cells cannot offer much protection and their chief function is to allow materials to pass through via diffusion and osmosis. Simple squamous epithelium is found in the walls of lung alveoli, blood capillaries, and bowman's capsules of nephrons.
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I will cut the chips so that they are all the same length, width and depth. Total volume of solution The more of the solution, the quicker the experiment will perform, again proving unfair. If we had used different amounts of solution in the test tubes it would affect the rate of osmosis taking place. If you used more solution for one than another it would dramatically effect the results. The one in more solution would result in the potato chip being much longer than the chip in less, because there is a larger volume of water molecules in the larger volume of solution which means there will be more diffusing into the potato through the semi-permeable membrane.
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AN ACCOUNT OF NITROGENOUS EXCRETION IN MAMMALS Excretion is the disposal of the waste products of metabolism, example ammonia and urea from
The three compounds of importance in nitrogenous excretion of animals are ammonia, urea and uric acid. The amino group being removed from excess amino groups by the process of deamination forms them. Ammonia is an extremely soluble gas of lower molecular mass. It diffuses rapidly, even when dissolved in water. Ammonia is a highly toxic substance: organisms cannot tolerate an accumulation of ammonia. The toxicity of ammonia is reduced by dilution with relatively large amounts of water. Animals of fresh water habitats continuously take in water by osmosis, and so are able to excrete ammonia in dilute solution without suffering any dehydration.
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Excretion is the biological process by which an organism chemically separates waste products from its body. The waste products are then usually expelled from the body by elimination.
where it is filtered and then flows into the Bowman's capsule. This process is called glomerular filtration. The water, waste products, salt, glucose, and other chemicals that have been filtered out of the blood are known collectively as glomerular filtrate. The glomerular filtrate consists primarily of water, excess salts (primarily Na+ and K+), glucose, and a waste product of the body called urea. Urea is formed in the body to eliminate the very toxic ammonia products that are formed in the liver from amino acids.
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As there are 6 tubes I will have to calculate how many pieces I will need altogether. I think this would be a suitable calculation: 6x3=18. So, I will pierce out some pieces of a potato using a core, and then cut 18 slices with a knife. Then I will measure 10ml of each of the 6 sucrose solutions using a measuring cylinder. I will be given these different concentrated solutions in the general laboratory. I will label the tubes 1 to 6 and fill each one with 10ml of the solution.
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In this experiment, water can pass through the selectively permeable membrane, but not the sugar molecules, this is because the semi permeable membrane "imprisons" the sugar molecules, Because of the higher concentration of water molecules in the beaker, more water passes into the thistle funnel than out of it, and the level of sugar solution in the funnel rises as water diffuses through into it. (This is shown in the diagram below). Thistle Funnel 2nd level of sugar solution after several hrs Water in beaker 1st level of sugar solution Selectively Strong sugar solution permeable membrane The sugar solution is said to have a low water potential.
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Diffusion and Osmosis: Investigating the Processes of Diffusion and Osmosis in a Model Membrane System and the Effect of Solute Concentration on Water Potential as it Relates to Living Plant Tissues
This enabled us to explore what is meant by the term "water potential." We determined the water potential of potato cells by recording and observing the initial mass of four potato cylinders, and then put the cylinders into a sucrose solution that were left overnight. We recorded the final mass the following day, and calculated the mass difference and percent change in mass. The lab was successful because it taught us the principles of diffusion, osmosis, and water potential. Our observations and results concurred with what was expected to happen. Introduction All atoms and molecules have kinetic energy and are constantly in motion.
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However this may at some point peak and therefore cease to increase in speed. Fermentation is an anaerobic process and which turns glucose into ethanol and carbon dioxide. In this experiment the yeast will be carrying out respiring anaerobically and breaking down the glucose and giving ethanol and carbon dioxide as waste products. There are enzymes in the yeast to break the glucose down. My reason for my prediction is that the more glucose that is present in the yeast the more will be broken down and therefore more carbon dioxide and ethanol will be produced and at a faster rate.
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An investigation to see whether the concentration of Sucrose effects the amount of Carbon Dioxide released from the respiration in yeast cells
As can be seen above in the word equation the products of the fermentation are ethanol and carbon dioxide. The greater the concentration of the sucrose solution means that there are more sucrose molecules in the solution compared to a lower concentration of the sucrose solution of the same volume. This means that if the same amount of yeast was put into both solutions (of a greater and a lesser concentration) more carbon dioxide and ethanol will be produced from the solution with the greater concentration because the raw materials are the limiting factor. This means that if there is less sucrose in a solution less carbon dioxide will be produced because the yeast will run out of raw material to feed on earlier than a higher % concentration (provided the yeast quantity is the same).
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Turgid is when a cell is so filled with water it can't take any more in. an animal cell would eventually burst but because the plant cell has a cell wall it doesn't burst. Water diffuses into the vacuole and pushes out on the cell membrane and wall till the cell wall can't expand any longer. Osmosis diagram - this shows that the water goes from the where there is a lot of it to where there isn't as much.
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