Practice 2: ORGANIC COMPOSITIONS OF THE CELL
1.CARBOHYDRATES
1.2.3. Task2-Chemical Detection of Starch Duration (mins) | Color of spot | Explanation | 0 | | _ For amylose (the polysaccharides that is present most in the rice suspension), the core of the helix structure easily accommodate iodine which is in Lugol solution). This complex has the deep blue color._ However, the helix structure (to be more specificly, the glycosidic bonds) is easily wrecked by heat (boiling temperature of water) and acid treatment (HCl). Once the helix structure of homopolycarbohydrate is broken, the conformation with iodine no longer exists. It leads to the gradual loss of blue color. | 2 | | | 4 | | | 6 | | | 8 | | | 10 | | | | No color change | | Color of Lugol | | |

2.PROTEINS 2.2.2. Task-Qualitative Detection of Proteins Protein solution | Original color | After 10% NaOH | After 0.5% CuSO4 | 1. Egg albumin | Colorless | Colorless | Blue->purple | 2. Fresh cow milk | White | White | Blue->purple |
Explanation:
NaOH doesn’t react with Proteins, so there’s no change to the protein solutions
Next, when CuSO4 is added, it reacts with NaOH to form Cu(OH)2 (blue) first.
2NaOH+ CuSO4 -> Cu(OH)2(s) + Na2SO4
Then Cu(OH)2 and Proteins together form the complex, which gives out the purple color. This reaction (called biuret reaction) is useful to detect the protein presence.
(write down the chemical equation, GOOGLE) 3. LIPIDS 4. NUCLEIC ACIDS
4.1.General Introduction
Nucleic acids are large biological molecules essential for all known forms of life. They include DNA (deoxyribonucleic acid) and RNA(ribonucleic acid). Together with proteins, nucleic acids are the most important biological macromolecules; each is found in abundance in all living things, where they function in encoding, transmitting and expressing genetic information.
Nucleic acids are macromolecules that exist as polymers called polynucleotide, each polynucleotides consists of monomers called nucleotides. A nucleotide is itself composed of three parts:a nitrogenous bas , a five-carbon sugar (a pentose), and a phosphate group.

To build a nucleotide, there are two families of nitrogenous bases: pyrimidines and purines. A pyrimidine has a six-membered ring of carbon and nitrogen atoms. (The nitrogen atoms tend to take up H+ from solution, which explains why it's called a nitrogenous base.)The members of the pyrimidine family are cytosine(C), thymine(T), and uracil(U). Purines are larger, with a six-membered ring fused to a five-membered ring. The purines are adenine(A) and guanine(G). The specific pyrimidines and purines differ in the chemical Groups attached to the rings. Adenine, guanine, and cytosine Are found in both types of nucleic acid; thymine is found only in DNA and uracil only in RNA.

PRACTICAL 3: PHOTOSYNTHESIS AND TRANSPIRATION
1.PHOTOSYNTHESIS
1.2.2. Task1-Examination of Oxigen Formation Duration (days) | Level of Water (mm) | Explanation | 0 | 0 | _The level of water has been lower and lower by days. This is the result of forming gas inside the test tube. This gas is lighter than air because it appears in the top of the tube. _Moreover, if we make a experiment with burned match, the match will burn up again, which proves that this gas oxygen(write down the chemical equation of photosynthesis) | 2 | 0.4 | | 4 | 0.6 | | 6 | 1.1 | | 8 | 1.9 | | 10 | 2.6 | |

1.2.3.Task2-Examination of starch formation Leaf samples | Original color | After 1-week covering | After boiling | After Lugol | Sample 1 | Green | Nothing changes | Lighter-> colorless | Nothing change | Sample 2 | Green | Nothing changes | Lighter -> colorless | Dark blue |

Explanation: _The green of the leaf is made by the chlorophyll (a kind of pigment) which is in the chloroplast. Being covered a week doesn’t affect much the contribution of chloroplast so the leaf maintain its green color. _Starch is stored in the stroma of chloroplast after photosynthesis. After being boiled with water and ethanol, 2 outer layers of the leaf and double membrane of the chloroplast is broken down. Because chlorophyll dissolves in ethanol, the green color gradually disappears. _After Lugol, the part of the leaf which was covered (sample 1) shows no change because it couldn’t do photosynthesis. The other part (sample 2) gives out the dark blue color because of the starch that was made in photosynthesis.

2. TRANSPIRATION Absorbent paper | Color observation | Explanation | Original color | White | _The Co2+ ion can form the complex with absorbed water and change its original color (blue) to the color of its conjugate with water (pink)(write down the equation)_That’s why the white paper becomes pink after soaking in CoCl2 solution and gives out blue color after drying.After sticked on the surface of the leaf, the paper’s color changes gradually from blue to pink because of the water that is produced by the transpiration of the leaf._ After the paper gets pink, the longer the time is, the more water is produced and the lighter the pink color is. | After soaking in 3% CoCl2 solution | Pink | | After drying | Blue | | Duration after applying on leaf (mins) | Color observation | | 0 | blue | | 5 | blue | | 10 | bluish | | 15 | bluish | | 30 | pink | | 60 | pinkish | |

PRACTICE 4: ENZYMES 1.2.2.Procedure Temp | Marked tubes | Color Observation | Explanation | | | Before Lugol | After Lugol | | 4 | 4-S | Colorless | Dark brown | _Amylase acts on alpha-glycosidic bonds in starch, separates them into single glucose molecules._Besides, Lugol solution reacts with starch and forms blue complex_Therefore, the color in the “S” test tube is darker than that in the “E” test tube. (the original color of Lugol solution is brow) _Enzyme (amylose) is less active and even inactivated in a temperature which is too high or low. in this case, they are 4 celsius degrees and boiling temperature of water. The starch is not splitted into glucose molecules. Hence, the Lugol solution reacts with starch and create the dark blue color | | 4-E | Colorless | Dark brown | | RT | RT-S | Colorless | Dark brown | | | RT-E | Colorless | Yellow brown | | 50 | 50-S | Colorless | Dark brown | | | 50-E | Colorless | Yellow brown | | 100 | 100-S | Colorless | Dark blue | | | 100-E | Colorless | Dark blue | |

2.PROTEAS 2.2.2. Procedure Temp. | Shape Change | Explanation | | At the beginning | After 2 days | | RT | A small piece | A part of white egg disolves | _In the suspension collected from pineapple, there is proteas which is a kind of enzyme that can cut peptide bonds in proteins._As we know, enzymes are inactivated in high temperature (in this case, 100 celsius degrees). Proteas can’t work properly in the “100” test tube, so nothing happens in this tube._In the other side, proteas work very well in room temperature. Therefore, the piece of white egg is “dissolved” because of the breaking of peptide bonds. | 100 | A small piece | Nothing change | |

3.CATALASE 3.2.2. Procedure

Temp. | Time recorded | Explanation | 4 | 2 mins 15s | _Enzymes can’t work properly in extreme temperatures (4 C and 100 C) and but in the right ones (50 and RT)._In addition, catalase can convert H2O2 into water and oxygen, which produces a force that pushes the round sheet upward. Hence, the faster oxygen is released, the stronger the papers fly to the surface of the test tubes. _As we can see, the tubes which are labeled “50” and “100” take much less time than the rest. | RT | 1 mins 30s | | 50 | 1mins 15s | | 100 | A very long time… | |