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MCV4U UNIT 4 – EXTENSIONS ASSESSMENT OF LEARNING CERTIFICATION SCRIPT 2026 QUESTIONS WITH SOLUTIONS GRADED A+

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MCV4U UNIT 4 – EXTENSIONS ASSESSMENT OF LEARNING CERTIFICATION SCRIPT 2026 QUESTIONS WITH SOLUTIONS GRADED A+

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MCV4U UNIT 4 – EXTENSIONS
ASSESSMENT OF LEARNING
CERTIFICATION SCRIPT 2026
QUESTIONS WITH SOLUTIONS
GRADED A+

◍ Water removed from the reactants joining two molecules together forming a
chemical bond.
Answer: Condensation
◍ The addition of water to the reactants to break a chemical bond between 2
molecules.
Answer: Hydrolysis
◍ explain binary fission in bacteria.
Answer: the circular DNA molecule gets replicated so therefore the
plasmids replicate, then the cytoplasm divides equally producing two
genetically identical daughter cells
◍ % increase in number of bacteria in 20 degrees versus 24 degrees celsius.
Answer: 10^5.2-10^4.1 divided by 10^4.1 multiplied by 100=1159%
◍ BD is a drug that inhibits ATP synthase in myobacteria by stopping
hydrogen ions movement. Explain how BD kills these bacteria..
Answer: No protons will be able to move through the ATP synthase
meaning there will be no energy to add inorganic phosphate to ADP, no
ATP for metabolic processes.
◍ Table 1 shows for statements and the names of four parts of a nephron- tick
the true statements for each part..

, Answer: ADH receptors- distal convoluted tube and collecting duct. Water
can be absorbed from filtrate- proximal convoluted tubule, distal convoluted
tubule and collecting duct. Blood is under high pressure- glomerulus.
Glucose is absorbed from filtrate- proximal convoluted tubule.
◍ Why is protein not present in urine?.
Answer: Plasma proteins are too large to filter through the basement
membrane.
◍ 1. Add Benedict's reagent. 2. Heat the solution in a water bath for 5 minutes
at 95 degrees Celsius. 3. Change from blue to brick red as CuO formed.
Answer: Test for Reducing Sugars (3)
◍ Evaluation question..
Answer: Small sample size, only done in rats not humans, only two
quantities, not a long time HOWEVER statistical test was carried out
showing that results were significant, rats and humans are both mammals.
◍ How many different genera of beetles are used in information provided..
Answer: 2
◍ Suggest two reasons why the first sample of beetles was captured during the
second week of august.
Answer: Beetles have enough time to emerge from soil, to allow beetles to
randomly distribute .
◍ Smaller units from which larger molecules are made.
Answer: Monomer
◍ 1. Add 2cm³ of food sample then add 2cm³ of dilute HCl and heat.2. Add
2cm³ of NaHCO3 then do test for reducing sugars..
Answer: Non-Reducing Sugars (2)
◍ Why beetles have to be identified before being marked.
Answer: all 3 beetles are green.
◍ Use the table provided to estimate population size of
C. graminis.

, Answer: 248+124=372. 405X372 divided by 124=1215
◍ State two precautions they would've taken each year to obtain comparable
data.
Answer: collect samples at same time of year, make sure marking isn't toxic.
◍ general structure of an amino acid.
Answer: nan
◍ Add drops of iodine to starch solution. Colour change to blue-black.
Answer: Test for Starch (1)
◍ 1. Mix Test solution with ethanol. 2. Shake for 1 minute then add water. 3.
Cloudy white emulsion.
Answer: Test for Lipids (3)
◍ 1. Obtain equal volumes of test solution and NaOH then add a few drops of
biuret solution (dilute copper (II) sulphate solution). 2. Colour change to
mauve/purple.
Answer: Test for Proteins (2)
◍ 1. Very high resolution. 2. Needs thin and dead specimen. 3. Artefacts can
occur (remnant left on object during prep, such as air bubbles)4. Uses
magnets to focus on specimen5. Uses electrons fired at sample.6. Is not in
colour.
Answer: Transmission Electron Microscope (5)
◍ 1. Inhibitor is similar in shape to substrate so it impermanently binds to the
active site. 2. Prevents ESC from forming, slowing rate.
Answer: Competitive inhibition (2)
◍ 1. Molecule will bind to allosteric site. 2. Binding causes a change in active
site. 3. Permanently preventing further ES
C. .
Answer: Non-competitive inhibition (3)
◍ statement describing protein structure.
Answer: sequence of amino acids joined by peptide bonds.

, ◍ Single B cell divides and produces 4000 plasma cells.each one of these
plasma cells produces 2000 molecules of antibody Q per SECON
D. A human male blood volume is 5.4 dm (54000cm)Calculate total number
of molecules of antibody Q per cm3 of blood 24 hours after production of
these plasma cells..
Answer: 24 hours is 86400 seconds. 2000 X 86400=1.728 X 10^81.728 X
10^8 X 4000=6.912 X 10^116.912 X 10^11 divided by 54000 is
1.2000001.28X10^8
◍ 1. DNA helicase breaks the hydrogen bonds between the base pairs 2. 2
single strands formed as the double helix "unzips". 3. Free DNA nucleotides
in the nucleoplasm bond to the complementary bases on the strand. 4. DNA
polymerase forms phosphodiester bonds between adjacent DNA nucleotides
via condensation reaction with the hydrolysis of ATP, forming the
phosphate backbone.
Answer: DNA Replication: Semiconservative (4)
◍ 3 reasons for scientists decision to do further investigations.
Answer: Only looking at immune response in lungs, day 9 results are
anomalous, only measuring the number of B cells.
◍ 1. ATP stores or releases only a small amount of energy at a time, so no
energy is wasted as heat.2. Small and soluble so easily transported3. Easily
broken down, so energy is released instantaneously4. Can be quickly
re-made5. Can make other molecules more reactive via phosphorylation6.
ATP can't pass out of cell, so the cell always has an immediate supply of
energy..
Answer: Describe 6 properties of ATP that make it a good energy source.
(6)
◍ suggest two reasons why hardy weinberg can't be used to estimate allele
frequencies of CLP.
Answer: HW cannot predict effect of environmental factors, mutations
occur.

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