BIOLOGY WITH HAMZA · MDCAT REVISION
Biological Molecules
Name: ____________________ Date: ______________
Recall first. Check second. Return to the ideas you missed.
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Learning targets
- Explain why water's polarity and hydrogen bonding matter to life.
- Compare carbohydrate and lipid structures with their storage roles.
- Relate levels of protein structure to biological function and denaturation.
Core notes
- Water's polarity and hydrogen bonding produce its life-supporting properties.
- Small differences in glucose bonding distinguish storage polysaccharides from cellulose.
- Hydrophobicity makes triglycerides compact stores and phospholipids membrane-formers.
- Protein sequence guides folding, and folding enables function.
Read, cover and redraw
Unequal charge distribution.
Water molecules attract one another.
Cohesion, heat capacity and solvent action.
Transport, temperature stability and metabolism.
Written recall
1. Why is glycogen well suited to rapid glucose release?
2. Why does denaturation not necessarily destroy the primary structure?
Apply the idea
1. A linear peptide contains 12 amino acid residues. How many peptide bonds are present, and how many water molecules are used in its complete hydrolysis?
- 11 bonds; 11 waters
- 12 bonds; 12 waters
- 11 bonds; 12 waters
- 12 bonds; 11 waters
My answer: _____ Reason: ____________________________________
2. In double-stranded DNA, adenine constitutes 28% of the bases. What percentage is cytosine?
- 28%
- 22%
- 44%
- 14%
My answer: _____ Reason: ____________________________________
3. Which interpretation best explains these results for a purified carbohydrate solution?
| Sample treatment | Observation |
|---|---|
| Before hydrolysis | Remains blue |
| After acid hydrolysis and neutralisation | Orange precipitate |
- Hydrolysis removes all carbohydrate
- It contains only free glucose before hydrolysis
- It contains a non-reducing carbohydrate that yields reducing products on hydrolysis
- It must be a protein
My answer: _____ Reason: ____________________________________
Answer key
Cover this section while attempting the questions.
Written recall
1. Its extensive branching creates many terminal points where enzymes can add or remove glucose units simultaneously.
2. It usually disrupts weak interactions responsible for secondary, tertiary or quaternary folding without hydrolysing the covalent peptide backbone.
Applied questions
1. A — A linear chain of n amino acids has n − 1 peptide bonds. Hydrolysis consumes one water per bond, so both values are 11.
2. B — A pairs with T, so A + T = 56%. The remaining 44% is split equally between G and C; cytosine is 22%.
3. C — A negative initial reducing-sugar test followed by a positive test after hydrolysis supports a non-reducing carbohydrate yielding reducing sugars. The data do not uniquely identify sucrose.