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Unit 1

Unit 1. Topic 1 (Statistical Analysis) Topic 3 (Chemistry of Life). Topic 1 – Statistical Analysis. 1.1.1 State that error bars are a graphical representation of the variability of data. 1.1.2 Calculate the mean and standard deviation of a set of values.

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Unit 1

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  1. Unit 1 Topic 1 (Statistical Analysis) Topic 3 (Chemistry of Life)

  2. Topic 1 – Statistical Analysis

  3. 1.1.1 State that error bars are a graphical representation of the variability of data. • 1.1.2 Calculate the mean and standard deviation of a set of values. • 1.1.3 State that the term standard deviation is used to summarize the spread of values around the mean. • Standard deviation- 68% of all values lie within +/- 1 standard deviation of the mean. 95% of values lie within +/- 2 standard deviations of the mean. • 1.1.4 Explain how the standard deviation is useful for comparing the means and the spread of data between two or more samples. • When comparing data between two sample sets, the closer the means and the standard deviations, the more likely the samples came from the same population. • 1.1.5 Deduce the significance of the difference between two sets of data using calculated values for t and the appropriate tables. • Criteria for t -test : Normal distribution Sample size of at least 10 The t -test can be used to compare two sets of data and measure the amount of overlap.(two-tailed, unpaired t test). If the value of t is > the critical value at .05, then there is a significant difference between the two sets of data, and the null hypothesis should be rejected. 7. • 1.1.6 Explain that the existence of a correlation does not establish that there is a causal relationship between two variables. • A student takes an exam during a lightning storm. The student fails the exam. Did the lightning cause the student to fail? No, not necessarily. Correlation does not imply causation.

  4. 1.1.1 State that error bars are a graphical representation of the variability of data. • Error bars can be used to show either the range of the data or the standard deviation.

  5. 1.1.2 Calculate the mean and standard deviation of a set of values. • Mean: measure of the central tendency (middle value) of the data. Check if distribution may be skewed and the mean may in fact not be the middle value. In excel we use =average (number 1, number2,..) • Standard Deviation (s): Measure of the spread of data around the mean. • Can be used either as a measure of variation within a data set or of the reliability of a measurement such as the mean.

  6. Arithmetic mean formula: • Standard deviation formula:

  7. *** You will be required to show standard deviation, mean and t-tests with labs conducted in the class. (Hint: Use Excel)

  8. 1.1.3 - State that the term standard deviation is used to summarize the spread of values around the mean. MEMORIZE!!

  9. Standard deviation is a measure of how spread out the data values are from the mean. • It is assumed that there is a normal distribution of values around the mean and that the data is not skewed to either end. • 68% of all the data values (measurements) in a sample can be found between the mean +/- 1 standard deviation. • 95% of all the data values in a sample can be found between the mean + 2SD and the mean -2SD

  10. 68% of the data lies between what range? • 95% of the data lies between what range?

  11. 1.1.4 - Explain how the standard deviation is useful for comparing the means and the spread of data between two or more samples. • small S.D.  narrow variation (less error/ less uncertainty) • larger S.D  wider variation (more error/ more uncertainty)

  12. Overlap of error bar  the difference is not statistically significant. • Sample A and Sample B are not significantly different. • No overlap  we CANNOT conclude that they are statistically different. • At this stage the student should proceed to a t-test to determine any statistically significant difference.

  13. 1.1.5 - Deduce the significance of the difference between two sets of data using calculated values for t and the appropriate tables. • If you carry out a statistical significance test, such as the t-test, the result is a P value, where P is the probability that there is no difference between the two samples. • A. When there is no difference between the two samples: • A small difference in the results gives a higher P value, which suggests that there is no true difference between the two samples • By convention, if P > 0.05 you can conclude that the result is not significant (the two samples are not significantly different). • B. When there is a difference between the two samples: • A larger difference in results gives a lower P value, which makes you suspect there is a true difference (assuming you have a good sample size). • By convention, if P < 0.05 you say the result is statistically significant. • If P < 0.01 you say the result is highly significant and you can be more confident you have found a true effect. • As always with statistical conclusions, you could be wrong! It is possible there really is no effect, and you had the bad luck to get sets of results that suggests a difference or not, where there is none.

  14. Drawing conclusions: • 1. State the null hypothesis and the alternative hypothesis based on your research question.Null Hypothesis: 'There is no significant difference between the height of shells in sample A and sample B.'Alternative Hypothesis: 'There is a significant difference between the height of shells in sample A and sample B'. • 2. Set the critical P level at P= 0.05 (5%) • 3. Write the decision rule for rejecting the null hypothesis. • If P  > 5% then the two sets are the same (i.e. accept the null hypothesis). • If P  < 5% then the two sets are different (i.e. reject the null hypothesis). • 4. Write a summary statement based on the decision. • The null hypothesis is rejected since calculated P = 0.003 < P = 0.05 two-tailed test • 5. Write a statement of results in standard English which includes the hypothesis • There is a significant difference between the height of shells in sample A and sample B.

  15. 1.1.6 - Explain that the existence of a correlation does not establish that there is a causal relationship between two variables. • Positive correlation • Both variables increase OR decrease • Negative correlation • One variable increases and other decreases OR vice versa • Example: • Cigarette consumption and lung cancer deaths • Predict a correlation

  16. Are there other possible variables that could influence this relationship that are not shown on the graph?

  17. IF statistical analysis of data indicates a correlation between the independent and dependent variable this does NOT prove any causation (Cause and effect relationship). Only further investigation will reveal the causal effect between the two variables. • Correlation does not imply causation. Here are some unusual examples of correlation but not causation's • Ice cream sales and the number of shark attacks on swimmers are correlated. • Skirt lengths and stock prices are highly correlated (as stock prices go up, skirt lengths get shorter). • The number of cavities in elementary school children and vocabulary size have a strong positive correlation. • Clearly there is no real interaction between the factors involved simply a coincidence of the data. • Once a correlation between two factors has been established from experimental data  to continue the research to determine what the causal relationship might be, if any.

  18. Topic 3 – Chemistry of Life

  19. 3.1 - Elements • 3.1.1 Frequently occurring elements in living things • 3.1.2 Additional elements in living thing. • 3.1.3 Role of elements in living things • 3.1.4 Structure of water • 3.1.5 Properties of water • 3.1.6 Functions of water in living things

  20. 3.1.1 State the most frequently occurring chemical elements in living things. • Carbon - Forms 4 covalent bonds Ex. CH4 • Hydrogen – Forms 1 covalent bond, Ex. : H2 • Oxygen – Forms 2 covalent bonds, Ex. CO2 • Nitrogen – Forms 3 covalent bonds, Ex. NH3

  21. 3.1.3 State one role for each of the elements named in 3.1.2 3.1.2 State that a variety of other elements are needed by living organisms • Sulfur • Calcium • Phosphorus • Potassium & Sodium • Iron • Sulfur: essential element in variable group of some amino acids (therefore proteins) • Calcium: essential element in bones, teeth, shells, nerve function • Phosphorus: essential element in nucleotides, including ATP • Sodium and potassium: essential ion in neuron membrane potential, required for nerve impulse transmission • Iron: essential element in heme group of hemoglobin, oxygen transport molecule

  22. 3.1.4 Draw and label a diagram showing the structure of water molecules to show their polarity and hydrogen bond formation

  23. Water polarity https://www.youtube.com/watch?v=ASLUY2U1M-8

  24. 3.1.5 Outline the thermal, cohesive and solvent properties of water Thermal - hydrogen bonds between polar water molecules cause water to resist change • high specific heat (energy required to change water temperature) • high heat of vaporization (energy required to boil water) • high heat of fusion (loss of energy required to freeze water) • thus, water produces a stable environment for aquatic organisms

  25. Cohesive • hydrogen bonds between polar water molecules cause them to cohere • allowing for transpiration in plants moving water against gravity • surface tension between cohering water molecules • allowing for animals such as water striders to walk over the surface of ponds even though they are denser than water

  26. Solvent • the polarity of water attracts, or dissolves, any other polar or charged particles by forming hydrogen bonds with them • proteins, glucose, or ions, such as sodium or calcium are all soluble

  27. 3.1.6 Explain the relationship between the properties of water and its uses in living organisms Coolant • hydrogen bonds between polar water molecules cause water to resist change • high heat of vaporization (energy required to change liquid water to vapor) because hydrogen bonds must be broken • thus, evaporation of water from plant leaves (transpiration) or from human skin (sweat) removes heat, acting as a coolant

  28. Medium for metabolic reactions • cytoplasm is primarily water, providing a polar medium in which other polar or charged molecules dissolve • many enzymes are globular proteins that are water soluble so they dissolve in cytoplasm where they control metabolic reactions

  29. Transport medium • hydrogen bonds between polar water molecules cause them to cohere • as water is lost by transpiration/evaporation from plant leaves, hydrogen bonds between adjacent water molecules pull water up columns of xylem • thus, plants move water against gravity, by as much as 100 meters

  30. 3.2 – Basic Molecules • 3.2.1 Organic and Inorganic compounds • 3.2.2 Common organic molecules • 3.3.3 Examples of carbohydrates • 3.3.4 Functions of carbohydrates • 3.3.5 Organic polymers, condensation and hydrolysis reactions (general). • 1) Formation of a disaccharide • 2) Formation of polysaccharide • 3) Formation of a dipeptide and polypeptide • 4) Formation of a triglyceride • 3.3.6 Functions of lipids • 3.3.7 Energy in carbohydrates and lipids

  31. 3.2.1 Distinguish between organic and inorganic compounds • Organic: compounds containing carbon that are found in living organisms • exceptions: carbonates (e.g. CaCO3), hydrogen carbonates (e.g. HCO3), and oxides of carbon (e.g. CO, CO2) • all other compounds are regarded as inorganic

  32. 3.2.2 Identify amino acids, glucose, ribose and fatty acids from diagrams showing their structure Amino Acid • There are 20 common amino acids • Amino acids are monomers which combine to form the larger polypeptides. In turn polypeptides combine to form proteins. • Proteins molecules are the basis of enzymes and many cellular and extra cellular components. • Each of the common amino acids has the same structure as the one shown except that the R group is different. • Amino acids are soluble

  33. Glucose: C6H12O6 Hexose sugar (six carbons) most commonly found in this ring structure. • product of photosynthesis or the substrate molecule for respiration. • polymer as starch, glycogen or cellulose. • All bonds are covalent. • Is a reducing sugar and will give positive (Brick red) precipitate in a Benedicts test. • metabolically active compound • soluble and has osmotic effects when in solution

  34. Ribose: Pentose (5 carbon sugar). • Part of one the important organic molecules in photosynthesis, ribulosebisphosphate. (RUBP) • A modified version of ribose, deoxyribose is perhaps best known for its role in Deoxyribonucleic acid or DNA where it forms part of the sugar phosphate backbone. The chemical properties of deoxyribose are very different from the properties of ribulose • Both Ribose and Glucose will attract water molecules (hydrogen bonding) to form solutions.

  35. Fatty Acid • Basis of triglycerides and many other types of lipid. These molecules are also the basis of the phospholipid molecules that form the bilayer of the cell membrane. Saturated fatty acid = (no double bonds) • Chain is the formed from a series of covalently bonded carbons saturated with hydrogens. • The chain is non-polar and hydrophobic • The carbonyl group is polar making this ends of the molecule hydrophilic. • Animals fats have saturated fatty acids which are straight molecules and very compact. This is gives them a higher melting point than the plant oils

  36. Unsaturated Fatty Acid = double bonds • If there are many double bonds the fatty acid is known as polyunsaturated. • Plant oils have unsaturated and polyunsaturated fatty acid chains that tend to branch and make the molecule less dense and with a lower melting point.

  37. Micelle • In water fatty acid molecules arrange themselves into spheres called micelles. • The polar carbonyl groups on the outside in contact with water molecules. • The non-polar tail sections are in the centre away from water. • This is an important aspect of fat digestion and membrane structure.

  38. 3.2.3 List three examples each of monosaccharides, disaccharides and polysaccharides

  39. 3.2.4 State one function of glucose, lactose and glycogen in animals, and of fructose, sucrose and cellulose in plants

  40. 3.2.5 Outline the role of condensation and hydrolysis in the relationships between monosaccharides, disaccharides and polysaccharides; between fatty acids, glycerol and triglycerides; and between amino acids and polypeptides • (1) Dimers a) Two monomers are bonded together to form a dimer. b) Water (H + OH) are removed to form water. c) The dimer can be split by hydrolysis but needs water • (2)Polymerisation a) In this example six monomers are joined together b) Polymers normally form more complex shapes than suggested in this model c) The polymer can be 'digested' back to monomers by hydrolysis reaction

  41. Formation of a disaccharide • a) Two molecule of glucose will polymerise to form maltose • b) The condensation reaction will take place between C1 of the first glucose and C4 of the second glucose. • c) A condensation reaction takes place between the glucose 1 (-OH on C1) and Glucose(-H on C4). • d) The bond formed is a covalent bond between C1 -O-C4 , called a 1, 4 glycosidic bond. • e) The disaccharide molecule formed is called Maltose. • f) Hydrolysis; The diagram can be reversed so that the disaccharide can be split into two glucose monosaccharides. • g) Hydrolysis is the type of reaction catalysed by the digestive enzymes.

  42. Polysaccharide Formation • The chain of glucose molecules represent the polysaccharide formed by many glucose monomers joining together to form this polysaccharide called amylose. • The molecule to the right represents the helical structure of the polypeptide, amylose. • Amylose is a polymer of glucose. • Intramolecular hydrogen bonding causes the chain molecule to twist into a helical shape. • Amylose is one of two molecules found in starch, the other being a branching polymer of glucose called amylopectin.

  43. Starch: Starch is composed of two polysaccharides, Amylose and amylopectin • Starch is metabolically un-reactive and insoluble and hence an excellent storage carbohydrate.

  44. Formation of a dipeptide and a polypeptide • Two amino acid monomers of glycine align • The peptide bond can form between the carboxyl group of the first amino acid and the amino group of the second amino acid. • H-OH or water is removed • Dipeptide is formed (naming system not required) with the characteristic -C-N- bond between the two monomers.

  45. Animation - http://www.cengage.com/biology/discipline_content/animations/reaction_types.html

  46. Polypeptide: • Polypeptide chains do not remain as linear (straight) chains. They fold up into complex yet specific shapes of the protein. • Shape - determined by intra-molecular hydrogen bonding and some covalent bonding between R groups (-S-S-, disulphide bridges).

  47. Protein formation -https://www.youtube.com/watch?v=iaHHgEoa2c8

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