Pearson Edexcel International A Level Biology: Decomposition, PCR, Electrophoresis and Forensic Biology (YBI11)

This guide follows the Pearson Edexcel International Advanced Subsidiary/Advanced Level Biology specification (YBI11), Issue 2, and covers statements 6.16 to 6.20 at the end of Topic 6, Microbiology, Immunity and Forensics. Topic 6 is in Unit 4, so it belongs to the International A2 (IA2) part of the course. There are no tiers, and each statement below is covered in full.

Find the rest of the course on the Edexcel A Level Biology hub and tick off progress on the printable checklist. Companion pages: revision notes and practice questions. To find weak spots first, try the free 10-minute diagnostics. For wider exam technique see the Edexcel IAL Biology exam preparation guide.

What this unit covers

Statement	What you must be able to do
6.16	Describe the role of microorganisms in the decomposition of organic matter and the recycling of carbon
6.17	Explain how DNA is amplified by the polymerase chain reaction (PCR)
6.18	Explain how gel electrophoresis separates DNA fragments of different length
6.19	Explain how DNA profiling identifies individuals and shows genetic relationships in plants and animals
6.20	Explain how the time of death of a mammal is estimated from decomposition, succession, forensic entomology, body temperature and muscle contraction

6.16 Decomposition and the recycling of carbon

Dead plants, dead animals and faeces contain organic molecules: carbohydrates, proteins, lipids and nucleic acids. Decomposition is the breakdown of these molecules. The main decomposers are saprotrophs, which are bacteria and fungi that feed on dead organic matter.

How a saprotroph feeds:

1. It secretes digestive enzymes onto the dead material (extracellular digestion). Examples are proteases, lipases and carbohydrases such as cellulase.
2. The enzymes hydrolyse large molecules into small soluble ones such as amino acids, fatty acids and sugars.
3. The microorganism absorbs the products across its cell surface. A fungus has thread-like hyphae that give a large surface area for secretion and absorption.
4. It uses some products for growth, so carbon is built into new microbial biomass, and respires others. Respiration releases carbon dioxide.

Detritivores such as earthworms and woodlice eat dead material and break it into small pieces. This increases the surface area for saprotrophs but they are not microorganisms.

Recycling carbon. Photosynthesis in producers fixes carbon dioxide into organic compounds. Feeding passes the carbon along food chains. Respiration by producers, consumers and saprotrophs returns carbon dioxide to the atmosphere, where it can be fixed again. Combustion also returns carbon dioxide, and fossil fuels form where dead material was buried and not fully decomposed.

Decomposition is faster when it is warm (more enzyme activity, within limits), moist, well supplied with oxygen (for aerobic respiration) and near neutral pH. It is slow in cold, dry, acidic or waterlogged conditions, which is why peat builds up.

Worked example. Litter bags hold 25.0 g of dry leaves. After 70 days the leaves weigh 17.5 g.

• Mass lost = 25.0 − 17.5 = 7.5 g
• Percentage loss = 7.5 / 25.0 × 100 = 30.0%
• Mean rate = 7.5 / 70 = 0.11 g per day (2 s.f.)

The lost mass is mostly carbon released as carbon dioxide by respiration, plus soluble products washed out.

6.17 Amplifying DNA: the polymerase chain reaction

PCR makes millions of copies of a chosen DNA sequence in a tube (in vitro). It needs no living cells and works from a tiny sample.

The tube contains:

• template DNA containing the target sequence
• two primers, short single-stranded DNA sequences complementary to the ends of the target, which mark where copying starts
• free DNA nucleotides to build new strands
• Taq polymerase, a DNA polymerase from the heat-tolerant bacterium Thermus aquaticus, which is not denatured at the high temperature of the first stage
• a buffer, with magnesium ions that the enzyme needs

A thermocycler repeats a three-stage cycle. Temperatures are typical values:

Stage	Temperature	What happens
Denaturation	about 95 °C	Hydrogen bonds between the bases break, giving single strands
Annealing	about 50 to 65 °C	Primers form hydrogen bonds with their complementary sequences
Extension	about 72 °C	The polymerase extends the 3′ end of each primer one nucleotide at a time, building new complementary strands

Each cycle roughly doubles the number of target copies, so after n cycles there are about 2ⁿ times as many (an idealised figure: the first cycles also make longer products that are not the target length).

Worked example. A sample has 4 target molecules. Each cycle takes 1 minute 30 seconds and the reaction runs for 25 cycles.

• Copies = 4 × 2²⁵ = 4 × 33 554 432 = 134 217 728
• Time = 25 × 1.5 = 37.5 minutes

For comparison, one molecule after 30 cycles gives 2³⁰ = 1 073 741 824 copies.

Why it matters: forensic samples are often tiny, and PCR gives enough DNA to profile. Contamination is the main risk, because a single stray cell’s DNA is also amplified.

6.18 Gel electrophoresis

Gel electrophoresis separates DNA fragments by length. Fragments may come from PCR or from DNA cut with restriction enzymes, which cut at specific base sequences.

Method:

1. Set an agarose gel in a tank and cover it with buffer, which conducts current.
2. Load each DNA sample into a well at the negative end. Load a DNA ladder (fragments of known length) in one well.
3. Switch on the current. DNA is negatively charged because of its phosphate groups, so fragments move towards the positive electrode.
4. The gel is a mesh. Shorter fragments move faster and travel further; longer fragments are held back.
5. Switch off before fragments run off the gel. Make the bands visible with a DNA stain viewed under UV light, or with fluorescent or radioactive labels.

Each band is a group of fragments of the same length. Compare band positions with the ladder to estimate sizes. Distance moved falls in a roughly straight line as the logarithm of fragment length rises, so a graph of log10 length against distance gives a straight line.

Worked example. A ladder gives 2000 base pairs (bp) at 12 mm, 1000 bp at 20 mm, 500 bp at 28 mm and 250 bp at 36 mm. Every 8 mm of travel halves the length. A fragment ran 24 mm.

• 24 mm is 4 mm past the 1000 bp band, which is half of one 8 mm halving step.
• Length = 1000 × 2^(−4/8) = 1000 × 0.707 = 707, so about 700 bp.
• A second fragment of 1000 bp (20 mm) with this one fits a 1700 bp piece of DNA cut once, since 1000 + 700 = 1700.

6.19 DNA profiling

Most human DNA does not code for protein. Some non-coding regions contain short tandem repeats (STRs), where a short base sequence is repeated end to end. The number of repeats at a locus varies between people. You inherit one allele for each locus from each parent, so everyone has two repeat numbers per locus (the same number twice if homozygous).

Method:

1. Collect a sample (blood, saliva, hair root, tissue) and extract DNA.
2. Amplify the chosen STR loci by PCR, using primers that flank each locus. Labels on the primers make the products fluorescent.
3. Separate the products by electrophoresis. Fragments with more repeats are longer and travel less far.
4. Compare the pattern of bands (the DNA profile) between samples.

Identification. A profile from a crime scene is compared with profiles from suspects. Different bands exclude a person. A full match supports identity, and the more loci compared, the less likely an unrelated person matches by chance. Identical twins share a profile.

Genetic relationships. In a child’s profile, each band must be present in the mother or the father. Bands not in the mother must come from the father.

Worked example (paternity). Repeat numbers at three loci:

Locus	Mother	Child	Man 1	Man 2
1	8, 11	8, 12	12, 13	10, 13
2	14, 17	14, 15	15, 16	15, 16
3	6, 9	9, 9	7, 9	6, 7

The child’s paternal alleles are 12 at locus 1, 15 at locus 2 and 9 at locus 3. Man 1 has all three, so he can be the father. Man 2 lacks 12 and 9, so he is excluded.

If the frequencies of the matching repeat patterns in the population are 0.12, 0.09, 0.15 and 0.06 at four loci, multiplying gives 0.0000972, about 1 in 10 000. This shows why using many loci makes a chance match unlikely. This calculation goes beyond the statement but shows how strong the evidence is.

Plants and animals. Profiling works for any organism. Uses include checking parentage in a breeding programme, choosing unrelated mates to limit inbreeding in captive animals, confirming the identity of a crop variety, and linking illegally traded plant or animal material to a source population.

6.20 Estimating time of death in a mammal

Pathologists use several indicators together because each is affected by conditions. The result is an estimated interval, not an exact time.

Extent of decomposition. After death, cells break down by self-digestion (autolysis) and bacteria from the gut spread and cause putrefaction. The body changes colour, swells with gas and softens. Warmth speeds this up; cold slows it.

Stage of succession. Different species colonise a body in a predictable order as conditions change. Blowflies arrive first, and later arrivals such as beetles feed on drier remains. The community present indicates how long the body has been there.

Forensic entomology. Blowflies lay eggs on a body. The life cycle is egg, larva (maggot, through successive moults), pupa, then adult. Larval length and stage show how long they have been feeding. Development is faster when warm, so use tables of development time for the species at the recorded temperature. The oldest stage present gives a minimum time since death, because insects arrive after death.

Body temperature. After death the body cools towards the surroundings. The time since death is estimated from the fall below normal body temperature. The cooling rate depends on air temperature, body size, clothing and wind, so it is a first estimate.

Muscle contraction (rigor mortis). After death ATP is no longer made. Myosin heads need ATP to detach from actin, so cross-bridges stay locked and muscles become rigid. Stiffness begins in small muscles, spreads, then fades as muscle proteins break down. It develops faster in warm conditions.

Worked example (body temperature). A body is found at 02:00 with a temperature of 28.5 °C. Assume it was 37.0 °C at death and cooled at 1.5 °C per hour (data given).

• Fall = 37.0 − 28.5 = 8.5 °C
• Time = 8.5 / 1.5 = 5.67 hours = 5 hours 40 minutes
• Estimated death = 02:00 − 5 h 40 min = about 20:20

Worked example (entomology). At 20 °C a blowfly species takes 1 day to hatch, 5 days as larvae and 8 days as pupae. The oldest insects on a body are pupae that are 3 days old.

• Time = 1 + 5 + 3 = 9 days at 20 °C
• The body has been dead at least 9 days. If the real temperature was higher, development was faster and the true time would be a little shorter.

Common errors

• Writing that Taq polymerase “works at high temperature because it is tough”. State that it is heat-stable, so it is not denatured during denaturation.
• Saying primers “join the strands”. They anneal to single strands to give the polymerase a start point.
• Saying DNA “is attracted to the negative electrode”. DNA is negative, so it moves to the positive electrode.
• Saying decomposers “give out” carbon. The carbon leaves as carbon dioxide from respiration.
• Giving a single indicator as an exact time of death.

Next steps

Use the revision notes for recall and the practice questions to test yourself.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Biology (YBI11), Specification Issue 2, February 2021, Pearson Education Limited: Topic 6, statements 6.16 to 6.20.