Section B

Compare gene-expression microarrays and SNP genotyping microarrays. Include:• What each measures• The sample preparation differences• Colour interpretation differences

Gene-expression microarrays measure mRNA levels. They show which genes move up or down in a sample

SNP genotyping microarrays measure fixed single-base changes in genomic DNA. They show which allele is present at each SNP site.

Gene-expression arrays start with RNA. The RNA is converted to cDNA and labelled with dyes.

SNP arrays start with genomic DNA. The DNA is fragmented and processed to allow allele calls.

In expression arrays, green shows lower expression in the treated sample. Red shows higher expression in the treated sample. Yellow shows equal expression.

In SNP arrays, colours link to the allele at the SNP site. Each dye marks a base, so the colour tells you which nucleotide is present.

Analyse the following scenario:

A microarray heat map shows consistent green signals for a set of 50 genes in a treated cell line compared to the control.

Explain:

(a) What this colour pattern suggests

(b) What experimental validation method should follow and why

(a) The colour pattern

Green means those genes show lower signal in the treated cells than in the control. This suggests those 50 genes are downregulated in response to the treatment

.(b) Experimental validation

qRT-PCR should be done. It gives precise Ct values and fold-change data. This confirms that the drop in signal is real and not due to array noise or normal variation.

The treatment may block a shared pathway that controls these genes. For example, a drug might inhibit a transcription factor that normally activates these 50 genes. Loss of that factor would lower their expression across the set.

You are comparing treated vs untreated cell lines using microarrays. The processed data show that signal intensity for a set of genes is consistently lower in treated samples.
Interpret this finding and describe two possible biological reasons for the observed under-expression.

The lower signal in the treated samples shows that these genes are under-expressed after treatment. Their mRNA levels drop compared to the untreated group.

Two possible biological reasons are:

• The treatment may reduce transcription. This can happen if the treatment blocks a transcription factor that activates these genes.

• The treatment may reduce mRNA stability. Faster degradation lowers mRNA levels and produces weaker signals on the array

Explain how primer self-annealing or hairpin formation can negatively impact a qRT-PCR experiment.

Primer self-annealing or hairpin formation lowers the amount of primer available for the target.

The primer binds to itself instead of the template.

This reduces binding at the correct site and weakens amplification.

It also produces unstable Ct values because the reaction becomes less efficient.

Why is it best practice to use multiple housekeeping genes when analysing gene expression?

Using multiple housekeeping genes gives stronger normalisation.

It gives a stable average Ct across samples.

This supports accurate ΔCT and ΔΔCT values and reduces error in expression analysis.

Explain why Ct values from different samples cannot be directly compared without normalisation.

Ct values cannot be compared directly because each sample may have different RNA input, different cDNA yield or different reaction efficiency.

Explain how alkylating agents induce mutations and how these can lead to GC→AT transitions.

Alkylating agents add extra chemical groups to bases in DNA. This changes how the base pairs during replication.

When guanine is alkylated, it can pair with thymine instead of cytosine.

During the next round of replication, the altered pairing produces a GC to AT change.

This makes the transition permanent in the daughter strand.

A mutation results from improper repair of a double-strand break. Describe how NHEJ can create insertions, deletions, or translocations.

NHEJ joins broken DNA ends without using a template.

The broken ends are often uneven, so enzymes trim or fill bases to allow joining.

This process adds or removes bases and produces small insertions or deletions.

If ends from two different chromosomes join, the repair links the wrong pieces together. This creates a translocation.

Compare the DNA damage caused by ionising radiation to that caused by UV radiation.

Ionising radiation breaks the DNA backbone. It produces single-strand breaks and double-strand breaks. These breaks disrupt large regions and can lead to major structural change.

UV radiation does not break the backbone. It links adjacent pyrimidines and forms dimers. These dimers block replication and raise error risk.

Explain slipped strand mispairing and describe how it leads to insertions or deletions in repeat-rich regions..

Slipped strand mispairing happens during replication when the template and new strand lose correct alignment.

If the new strand slips forward, an extra repeat loop forms and an insertion appears in the next round.

If the template strand slips, a section is skipped and a deletion appears. The shift becomes fixed after replication and produces a stable indel.