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Sunday, April 13, 2014

PEKA FORM 4 Physics

Aim of the experiment:
To investigate the relationship between the period of a simple pendulum with its length.

Hypothesis:
The longer the length of a simple pendulum, the longer the period of oscillation. OR
The greater the length of a simple pendulum, the greater the period of oscillation.


Variables:
Manipulated: The length of the pendulum, l 
Responding: The period of the pendulum, T 
Constant: The mass of the pendulum bob, gravitational acceleration

Apparatus/Materials:
Pendulum bob, length of thread about 100 cm long, retort stand, stopwatch



Setup:
Procedure:

1. The thread is tied to the pendulum bob. The other end of the thread is tied around the arm of the retort stand so that it can swing freely.
2. The length of the pendulum, l is measured to 80 cm as per the diagram.
With the thread taut and the bob at rest, the bob is lifted at a small amplitude (of not more than 10°). Ensure that the pendulum swings in a single plane.
3. The time for ten complete oscillations of the pendulum is measured using the stopwatch.
4. Step 3 is repeated, and the average of both readings are calculated.
5. The period of oscillation, T is calculated using the average reading divided by the number of oscillations, i.e. 10. ex: T = t/10
6. T is calculated by squaring the value of T.
7. Steps 1 to 6 are repeated using l = 70 cm, 60 cm, 50 cm, and 40 cm.
8. A graph T2 versus l is plotted.
Data:




Analysis :



Discussion:
The graph of T2 versus l shows a straight line passing through the origin. This means that the period of oscillation increases with the length of the pendulum, with T2 directly proportional to l. 

Conclusion:
The longer the length of the pendulum, the longer the period of oscillation. The hypothesis is proven valid.

Precaution : (suggestion)
1. The experiment was carried out in an enclosed room to avoid the influence of wind.
2. Displace the pendulum with small angle to make sure it will oscillates in a straight line.
3. Avoid parallax error by make sure the position of the eyes is perpendicular to the reading on the stopwatch

Saturday, April 12, 2014

Sample Question & Answer for Mid-Year Exam F.4 Biology 2014

1. Diagram shows the fluid mosaic model of a plasma membrane.


(a) Name the parts labelled Y and Z.
Y : Carrier protein Z: Pore protein// Channel protein
(b)(i) State the component of structure X.
It is composed of two layers of phospholipids
(ii) Explain the main function of X.
Acts as a barrier between the internal and external environment of the cell
// Allows only specific molecules to pass through it
// provide the structural basis for all cell membrane
(c) The plasma membrane is said to be semi-permeable. What is the meaning of ‘semi- permeable’?
A semi-permeable plasma membrane is a membrane that allows only certain substances to move freely across it.
(d) The concentration of ions inside root cells is up to 100 times greater than in the soil. Anyway, the ions are still transported into the cells by active transport.
(i) Define active transport.
Active transport is a movement of substances / molecules / ions against the concentration gradient/from low to high concentration across the plasma membrane with the help of carrier protein and energy / ATP
(i) Explain what will happen to the uptake of the ions by root cells if the roots are immersed in a solution containing metabolic poisons such as cyanide.
P1 – there is no uptake of ions by root cells
P2 – metabolic poisons kill/ damaged the (root) cells
P3 – no energy/ ATP is produced
P4 – active transport does not occur
(e) Using the mechanism of movement of a substance across the plasma membrane through osmosis, explain why food is preserved using concentrated sugar solution.
P1- The glucose with a high concentration acts as a hypertonic solution
P2 - causing water in food to move out through plasma membrane
P3 – This causes food to dehydrate
P4 – condition not favourable to microorganisms 
2. Diagram shows a plant cell. X, Y and Z are structures found in the cell.
(a) (i) Name the structures X and Y. X : Golgi Apparatus/body Y: Rough Endoplasmic Reticulum
(ii) State the function of Z. Z : site for cellular respiration // to generate / produce energy
(b) Explain the function of X and Y in the transportation of extracellular enzyme.
P1 : The nucleus / RNA instructs ribosomes to synthesized protein
P2 : The synthesized protein is transported in the Rough Endoplasmic reticulum / Y
P3 : to the transport vesicles
P4 : then the Golgi Apparatus / X packages /modifies /sorts / transports the synthesized proteins
P5 : to the secretory vesicles to be transported
P6 : out of the cell through the plasma membrane
All enzymes are protein. Enzymes are sensitive to temperature.
(c)(i) Explain why food is kept in the refrigerator?
P1 : Temperature in the refrigerator is very low
P3 : Rate of enzyme reaction decreases as temperature decreases
P3 : Enzymes are inactive at low temperature
(ii) A branded washing machine is provided with temperature regulator. A housewife uses the detergent containing enzyme at 40C to wash the clothes. By using the information given, explain why?
P1 : The optimum temperature for enzyme reaction is about 40oC.
P2 : Low temperature makes the enzyme inactive
P3 : High temperature denatures the enzyme
(d) Figure below shows the structure of an enzyme and three substrates P, Q and R.
Based on the figure above, complete the schematic diagram below to show the mechanism of enzyme action on a suitable substrate.

3. A young spinach stem was cut into two longitudinal halves. Each was cut lengthwise again into two. This was repeated with another piece of spinach stem. The strips were of equal lengths and thickness. Two of these strips were placed into different sucrose solutions labelled A, B and C. The strips were observed 30 minutes later as shown below.
(a) Which of the above solution is :
(i) isotonic to the cell sap of the spinach? Solution B
(ii) hypotonic to the cell sap of the spinach? Solution A
(iii) hypertonic to the cell sap of the spinach? Solution C

(b) Explain why the strip placed in solution A curve outwards.
Solution A is hypotonic to the cell sap; water moves into the vacuole/cut region by osmosis, the cuticle restricts intake of water from epidermal cells, parenchyma cells/cut region expands and become turgid, strip curves outwards.
(c) Explain why the strip placed in solution C curve inwards.
Solution C is hypertonic to the cell sap; water moves out of the vacuole/cut region by osmosis, the cuticle restricts loss of water from epidermal cells, parenchyma cells/cut region shrinks and become flaccid, strip curves inwards.
(d) Explain why the strip placed in solution B remained unchanged.
Solution B is of the same concentration as the cell sap of strips in B; movement of water in and out of cell is the same. Thus strip in B remain unchanged.
(e) If the epidermis is removed from the strip in solution A, would you expect any curvature? Give your reason.
No; all parenchyma cells become equally turgid as water moves in equally at all sides and the length increases equally on all sides
4. An experiment was done to study the effect of pH on the activity of amylase. Three test tubes labeled P, Q and R were filled with the same amount of amylase.
The solution in P was neutral.
The solution in Q was acidic.
The solution in R was alkaline.
The same amount of starch was put into each test tube and the contents were stirred. Every two minutes a drop of the mixture from each test tube was dropped into a drop of iodine solution on a white tile. Table 1 shows the results of the experiment.
(a) P = 8 MIN Q = 10 MIN R = 14 MIN
(b)(i) State two different observations made from the Table 1.
Observation 1: Time taken for complete hydrolysis of starch in test tube P is 8 minutes
Observation 2: Time taken for complete hydrolysis of starch in test tube R is 14 minutes
(ii) State the inferences from the observation in 1(b)(i).
Inference from observation 1: Time taken for complete hydrolysis of starch in test tube P is fastest because amylase works best in a neutral medium
Inference from observation 2: Time taken for complete hydrolysis of starch in test tube R is slowest because amylase is inactive in an alkaline medium
(c) Complete Table 2 based on this experiment.
Variable
Method to handle the variable
Manipulated variable
pH

Use different medium of solution, acidic, alkaline and neutral
Responding variable
Time taken for complete hydrolysis of starch
Measure and record the time taken for complete hydrolysis of starch using a stopwatch
Measure and record the time taken for the dark blue colour to disappear using a stopwatch
Constant variable
Volume of amylase

Volume of starch
Temperature
Fix the volume of amylase at 1ml throughout the experiment
Fix the volume of starch 5ml throughout the experiment
Fix the temperature to be the same throughout the experiment

(d) State the hypothesis for this experiment.
The enzyme amylase hydrolyses starch at optimal pH which is neutral.
(h) State the operational definition for enzyme.
Enzyme/amylase hydrolysed/digest/act on starch and the time taken for the dark blue colour to disappear is affected by the pH of medium.
5. The figure shows the organelles involved in the production of extracellular enzymes.
(i) Extracellular enzyme is produced in a cell, then packed and secreted from the cell. It catalyses its reaction outside the cell. An example is amylase/pepsin/trypsin.
(ii) The instruction for making the extracellular enzyme is transcribed from the deoxyribonucleic acid (DNA) to ribonucleic acid (RNA) in the nucleus.
The RNA then leaves the nucleus through the nuclear pore and attaches itself to the ribosome located on the endoplasmic reticulum.
When the synthesis of the enzymes is completed, it is encapsulated in a transport vesicle which fuses with the golgi body.
In the golgi body, the enzyme is further modified before being packed in a secretory vesicle.
The secretory vesicle transports the enzyme to the plasma membrane, where it fuses with it and the enzyme is released outside the cell.
(b) The explanation of enzyme action is known as the ‘lock and key hypothesis’.
The substrate molecule fits into the active site of the enzyme molecule.
The substrate is the ‘key’ that fits into the enzyme ‘lock’. The reaction is specific.
Various types of bonds such as hydrogen and ionic bonds hold the substrate in the active site forming the enzyme-substrate complex.
Once the complex is formed, the enzyme changes the substrate to its product.
The product leaves the active site. The enzyme is not altered by the reaction and it can be reused.